Do ordered fields and archimedian ordered fields have the same first-order theory?

Let us suppose that the first-order language of ordered fields has symbols for addition, subtraction, multiplication and order, and constant symbols for 0 and 1. An ordered field is said to be archimedian if for any positive elements $x<y$ there is a positive integer $n$ such that $nx>y$.

Question: Is the first-order theory of archimedian ordered fields identical to the first-order theory of ordered fields? The non-trivial part of this question is: Suppose a sentence holds in every archimedian ordered field. Does it hold in every ordered field?

Here is a statement which is true in archimedian fields but not in general ordered fields:

For all $0 < a < b$, there exists $x$ such that $a < x^2 < b$.

This is true in archimedian fields because every archimedean field contains $\mathbb{Q}$ and is contained in $\mathbb{R}$, and every interval in $\mathbb{R}_{+}$ contains the square of a rational.

Let $K$ be the field $\mathbb{R}(t)$, with $f(t)>g(t)$ if $f-g$ is positive for $t$ sufficiently large. I claim that there are no squares between $t$ and $t+1$. Proof: Define $\deg p(t)/q(t) = \deg p - \deg q$, where $p(t)$ and $q(t)$ are polynomials. Then every square has even degree, but every element between $t$ and $t+1$ has degree $1$.

I usually try to give some clue where I came up with these answers. Vague thought process here: I want a non-archimedean ordered field $R$ and a statement which is true in both $\mathbb{Q}$ and $\mathbb{R}$, but not in $R$. So $R$ better NOT be real closed, because real closed fields have the same first order theory as $\mathbb{R}$, and my statement had better USE that $R$ is not real closed. The axioms of real closed fields are all about existence, so a statement which fails in a non-real-closed field should like something like $\exists x : \cdots$. But it also has to be true in $\mathbb{Q}$, which has very few elements. How can I do that? Maybe something like $\forall y \exists x : \cdots$; then it would be true in $\mathbb{R}$ because there are lots of choices for $x$, and true in $\mathbb{Q}$ because there are few choices for $y$.

So I want an equation, dependent on a parameter, which is always solvable in $\mathbb{Q}$, but for a nontrivial reason. (If it would solvable for a trivial reason, that reason would probably hold in every ordered field.) How about $\forall y>0 \exists x_1, x_2, x_3, x_4 : y = x_1^2+x_2^2+x_3^2+x_4^2$? That's true in $\mathbb{Q}$, for nontrivial reasons, and trivially true in $\mathbb{R}$, but false in $\mathbb{R}(t)$ because sums of squares have even degreee. I don't know if it's true in other archimedean fields, though. Still, the idea of using the fact that squares are reasonably spread out in $\mathbb{Q}$, but only land in discrete lumps of even degree in $\mathbb{R}(t)$ sounds like a good one...