Free groups in some classes?

There is no "free solvable group", but there are "free solvable groups of length $n$" for any $n$. There are also free nilpotent groups of class $n$, free Burnside groups of exponent $k$; all groups that are abelian-by-exponent $k$; and more.

More precisely: a variety of groups is a class of groups that is closed under taking isomorphisms, subgroups, quotients, and arbitrary direct products (that is, if $G\in\mathcal{C}$ and $K\cong G$, then $K\in\mathcal{C}$; if $G\in\mathcal{C}$ and $H\lt G$, then $H\in \mathcal{C}$; if $G\in\mathcal{C}$ and $N\triangleleft G$, then $G/N\in\mathcal{C}$; and if $\{G_i\}$ is an arbitrary family of groups with $G_i\in\mathcal{C}$ for every $i$, then $\prod G_i\in\mathcal{C}$). Examples of varieties include "all abelian groups", "all groups of solvability length at most $n$"; "all nilpotent groups of class at most $c$"; "all groups such that every element is of exponent $k$"; and more.

If $\mathcal{C}$ is a variety of groups, and $G$ is any group, then there is a smallest normal subgroup $\mathcal{C}(G)$ of $G$ such that $G/\mathcal{C}(G)\in\mathcal{C}$ (this normal subgroup is, in fact, not merely normal, but fully invariant). For example, when $\mathcal{C}$ is the class of all abelian groups, $\mathcal{C}(G)$ is none other than the commutator subgroup of $G$. Proof of the existence of $\mathcal{C}(G)$: Let $\{N_i\}_{i\in I}$ be the collection of all normal subgroups of $G$ such that $G/N_i\in\mathcal{C}$; the class is nonempty, since $G/G$ is always in any variety. Then consider the obvious map from $G$ to $\prod G/N_i$; the product must be in $\mathcal{C}$ (it is a product of groups in $\mathcal{C}$), so the image of $G$ is in $\mathcal{C}$; the kernel of this map is the group $\mathcal{C}(G)$.

So if $X$ is any set, and $F(X)$ is the free group on $X$, then there is a smallest subgroup $\mathcal{C}(F(X))$ such that $F(X)/\mathcal{C}(F(X))\in\mathcal{C}$. The resulting quotient can be seen to have the same universal property relative to groups in $\mathcal{C}$ as $F(X)$ does relative to all groups. So we say that this quotient is the "relatively free $\mathcal{C}$-group on $X$." We call the original group $F(X)$ the "absolutely free group on $X$" to distinguish it.

Added: Just in case, remember that if $\mathcal{D}$ is a category in which the objects are sets and the arrows are maps of sets, then given a set $X$, the free $\mathcal{D}$-object on $X$ is an object $F(X)$ in $\mathcal{D}$, together with a set-theoretic inclusion $i\colon X\to F(X)$, such that for every object $D\in\mathcal{D}$ and every set-theoretic map $f\colon X\to D$ there exists a unique $\mathcal{D}$-morphism $\varphi\colon F(X)\to D$ such that $f=\varphi\circ i$. For the "absolutely free groups", the category is the category of all groups; for "free abelian groups", the category is the category of abelian groups. For "relatively free $\mathcal{C}$-group", the category is the category of all groups in $\mathcal{C}$.

The reason there is no "free solvable group" is that the class of all solvable groups is not a variety: it is not closed under arbitrary direct products. But if you put a bound on the solvability length then you do get a variety, and you can construct the corresponding "free solvable group of length $n$".

In addition you can take the pro-$\mathcal{C}$ approach that Matt E mentions.

If you are interested in learning more about varieties, these lie at the intersection of Universal Algebra and Group Theory. I would recommend Hanna Neumann's book Varieties of Groups (a bit old, but still the standard reference); also George Bergman's Universal Algebra notes (available here).

I think that you might have to bound the derived length to get precisely the notion that you want.

Notation: For any group $G$, let $G^{(i)}$ denote the $i$th iterated derived subgroup of $G$, so $G^{(0)} = G$ and $G^{(i+1)} = [G^{(i)},G^{(i)}]$). A group is solvable if $G^{(i)}$ is trivial for some $i$, and has derived length $d$ if $G^{(d)}$ is trivial, but $G^{(i)}$ is non-trivial for $i < d$.

Now let $F(X)$ be the free group on a non-empty set $X$ of generators, and consider $F(X)/F(X)^{(d)}$. On the one hand, this group is clearly solvable of derived length $d$. On the other hand, if $G$ is any solvable group of derived length $\leq d$, then giving a map of sets $X \to G$ will be equivalent to giving a map of groups $F(X)/F(X)^{(d)} \to G$. Thus $F(X)/F(X)^{(d)}$ is the free object on $X$ in the category of solvable groups of derived length $\leq d$.

One can make a similar construction if you consider e.g. the category of nilpotent groups of nilpotency class $\leq c$ (i.e. whose central series is of length $\leq c$).

If you really want to consider all solvable groups at once, then you can consider the projective limit $$\lim_{\leftarrow} F(X)/F(X)^{(d)},$$ the so-called pro-solvable completion of $F(X)$. Similarly, in the nilpotent setting, you can form the pro-nilpotent completion of $F(X)$. But now the "free object" is not literally in the category you want, but is instead is a projective limit of groups from that category. (This is often okay, though; is you search for "pro-nilpotent completion", you will see that pro-nilpotent completions of free groups come up a lot, e.g. in arithmetic geometry.)