Uniformly solvable families of polynomials
The answer is no. The following is a theorem of Hilbert:
Let $f(x_1, x_2, \ldots, x_n, t)$ be a polynomial with rational coefficients. Let $K_{gen}$ be the splitting field of $f$ over $\mathbb{Q}(x_1, \ldots, x_n)$, thought of as a polynomial in $t$. For $a_1$, ..., $a_n$ in $\mathbb{Q}$, let $K(a_1, \ldots, a_n)$ be the splitting field of $f(a_1, \ldots, a_n, t)$ over $\mathbb{Q}$. There are infinitely many $(a_1,\ldots, a_n)$ for which $\mathrm{Gal}(K_{gen}/\mathbb{Q}(x_1, \ldots, x_n)) \cong \mathrm{Gal}(K(a_1, \ldots, a_n)/\mathbb{Q})$.
Infinitely many is a very weak statement; the right statement is that this is true for all but a "thin set" of $n$-tuples of rational numbers. There are many theorems saying that thin sets are small. See this question for some more discussion in the case that $G$ is $S_n$.
The theorem which gets the famous name is Hilbert's Irreducibility Theorem: If $f(x_1, \ldots, x_n, t)$ is irreducible as a polynomial in $\mathbb{Q}(x_1, \ldots, x_n)[t]$ then, for all most all $(a_1, \ldots, a_n) \in \mathbb{Q}^n$, the polynomial $f(a_1, \ldots, a_n)(t)$ is irreducible as a polynomial in $\mathbb{Q}[t]$. The deduction of the result I stated from this one is pretty straightforward (it is the first application of Hilbert's irreducibility theorem mentioned in Wikipedia) but I had trouble finding you a reference that writes it out carefully.
If you are happy with the language of modern algebraic geometry, I recommend the first three chapters of these notes by Serre. If you are willing to read something lengthy, this Master's thesis looks like a beautiful winding tour through Hilbert's Irreducibility Theorem and its many applications
It appears to me the answer is essentially no, but there are some details to be checked. The main observation is this: under reasonable conditions, the Galois group of the generic fibre is canonically isomorphic to the Galois group of almost all fibres; thus, the generic fibre is solvable if and only if almost all fibres are solvable.
In more detail: Let $L$ be a finite Galois extension of $K$. By the primitive element theorem, $L$ is generated by a single element over $K$, say $x$. Let $f$ be the (monic) minimal polynomial of $x$ over $K$ and choose a subring $A \subseteq K$ satisfying these conditions:
- $A$ is a noetherian integral domain.
- $\operatorname{Frac} A = K$.
- $f \in A [x]$.
- The ring $B = A [x] / (f)$, considered as a subring of $L$, is closed under the action of $\mathrm{Gal}(L \mid K)$.
- $B$ is a projective $A$-module of rank $d = \deg f$.
This is certainly achievable in the case $K = \mathbb{Q} (t_1, \ldots, t_n)$ that you are considering: take $A$ to be the subring generated by $\mathbb{Q} [t_1, \ldots, t_n]$, the coefficients of $f$, and the coefficients of all the Galois conjugates of $x$; then $B$ will even be a free $A$-module of rank $d$.
Lemma 1. $\mathrm{Aut}(B \mid A)$, the automorphism group of $B$ as an $A$-algebra, is canonically isomorphic to $\mathrm{Gal}(L \mid K)$.
Proof. The hypothesis implies that each automorphism of $L \mid K$ restricts to an automorphism of $B \mid A$; since $B$ contains the generator of $L$, the restriction homomorphism $\mathrm{Gal}(L \mid K) \to \mathrm{Aut}(B \mid A)$ is injective. On the other hand, $\operatorname{Frac}$ is functorial, so any automorphism of $B \mid A$ extends to an automorphism of $L \mid K$, so the restriction homomorphism is in fact an isomorphism. ◼
Lemma 2. Let $\mathfrak{m}$ be a maximal ideal of $A$ and let $\kappa (\mathfrak{m}) = A / \mathfrak{m}$ be the residue field.
- $B (\mathfrak{m}) = B \otimes_A \kappa (\mathfrak{m})$ is a $\kappa (\mathfrak{m})$-algebra of dimension $d$.
- $B (\mathfrak{m})$ is a field if and only if $f$ is irreducible over $\kappa (\mathfrak{m})$.
- There is a canonical homomorphism $\mathrm{Aut}(B \mid A) \to \mathrm{Aut}(B (\mathfrak{m}) \mid \kappa (\mathfrak{m}))$, and it is injective if and only if distinct Galois conjugates of $x$ remain distinct in $B (\mathfrak{m})$.
Proof. All straightforward. ◼
Lemma 3. There exists a dense open subset $U \subseteq \operatorname{Spec} A$ such that, for all maximal ideals $\mathfrak{m}$ of $A$ that are in $U$, the Galois conjugates of $x$ remain distinct in $B (\mathfrak{m})$.
Proof. Two Galois conjugates of $x$, say $x'$ and $x''$, become equal in $B (\mathfrak{m})$ precisely if a certain finite set of equations holds in $B (\mathfrak{m})$; more precisely, there exist elements $h_1, \ldots, h_d$ in $B$ such that $x' \equiv x'' \pmod{\mathfrak{m}}$ if and only if $h_1 \equiv \cdots \equiv h_d \equiv 0 \pmod{\mathfrak{m}}$. This is because the $A_{\mathfrak{m}}$-module $B_{\mathfrak{m}}$ is free of rank $d$ (since $B$ is a projective $A$-module of rank $d$). Thus $x' \equiv x'' \pmod{\mathfrak{m}}$ if and only if $\mathfrak{m}$ is in a certain closed subset of $\operatorname{Spec} A$ of codimension $\ge 1$ (by Krull's Hauptidealsatz). The complement of such a closed subset is a dense open subset of $\operatorname{Spec} A$ (because $A$ is an integral domain), and the intersection of finitely many dense open subsets in $\operatorname{Spec} A$ is again a dense open subset, so we are done by the finiteness of $\mathrm{Gal}(K \mid L)$. ◼
Theorem. If there exists a dense open subset $U' \subseteq \operatorname{Spec} A$ such that, for all maximal ideals $\mathfrak{m}$ of $A$ in $U'$, $f$ is irreducible over $\kappa (\mathfrak{m})$, then there exists a dense open subset $U'' \subseteq \operatorname{Spec} A$ such that, for all maximal ideals $\mathfrak{m}$ of $A$ in $U''$, $B \otimes_A \kappa (\mathfrak{m})$ is a Galois extension of $\kappa (\mathfrak{m})$ and the canonical homomorphism $$\mathrm{Gal}(K \mid L) \cong \mathrm{Aut}(B \mid A) \to \mathrm{Gal}(B \otimes_A \kappa (\mathfrak{m}) \mid \kappa (\mathfrak{m}))$$ is an isomorphism.
Proof. Let $U$ be as in lemma 3, and take $U'' = U \cap U'$. This is a dense open subset, and by lemma 2, for any $\mathfrak{m}$ in $U''$, $B \otimes_A \kappa (\mathfrak{m})$ is a field and the canonical homomorphism is injective. However, Dedekind's lemma says that the size of $\mathrm{Aut}(B \otimes_A \kappa (\mathfrak{m}) \mid \kappa (\mathfrak{m}))$ is bounded above by the degree of $B \otimes_A \kappa (\mathfrak{m}) \mid \kappa (\mathfrak{m})$, so in fact the canonical homomorphism is a bijection, and moreover $B \otimes_A \kappa (\mathfrak{m}) \mid \kappa (\mathfrak{m})$ is a Galois extension. ◼
Remark. It's not clear to me when the main hypothesis of the above theorem holds. There is definitely something non-trivial to be checked: after all, if we take $K = \mathbb{Q}$ and $A = \mathbb{Z}$, then we have to contend with the fact that there exist polynomials that are reducible mod $p$ for every prime $p$ but yet irreducible over $\mathbb{Q}$. But perhaps in the case you are considering this cannot happen.