Every ordered field $\mathbb F$ contains $\mathbb Q$ in a canonical way. If the field is not Archimedean there exists an $x>n$ for all $n \in \mathbb N$. Since we are dealing with a field any polynomial expression $P(x)$ must lie in the field and be invertible unless it is the zero polynomial. So also $1/P(x)$ will lie in $\mathbb F$.
From the simple algebraic properties of inversion, addition and multiplication that hold in every field it appears that $\mathbb F$ must then contain the field of rational polynomials $\mathbb Q(x)$ as a subfield (although not necessarily in a canonical way).
My question whether the ordering on this subfield has to be the same ordering as the ordering of $\mathbb Q(x)$. I think this would be implied by $x^2>x$, which follows from $x>1$ and then multiplying with $x$. This should give $x^2 > \alpha x + \beta$, with $\alpha, \beta \in \mathbb Q$ and the ordering on the polynomial part is the same as the one you get on the polynomial ring $\mathbb Q[x]$.
But I don't see whether this implies that the embedding $\mathbb Q(x)$ is order preserving.