Let $R(x)$ denote the rational functions over the reals and $\overline{R(x)}$ its algebraic closure. Also let $P(X,Y)$ denote the set of partial functions from $X$ to $Y$, where partial functions means in simple terms a "function" for which not every term in the domain is mapped.

Lastly let $h$ be the canonical homomorphism $h:R(x)\rightarrow P(R,R)$ that is described by simply evaluating the rational function over the $R$.

My question is whether there exists a homomorphism $f:\overline{R(x)}\rightarrow P(R,C)$ such that $f|_{R(x)}=h$ and $f(\sqrt[n]{q})=\sqrt[n]{f(q)}$

It is clear or any element of $\overline{R(x)}$ that can be expressed through the field operations of the rational functions and roots we have an obvious corresponding partial function associated with it (simply evaluate the expression as you would in an grade school algebra class). However, there are many elements of $\overline{R(x)}$ that are not expressible through roots and arithmetic operations and so this questions boils down to is there meaningful and consistent way of "evaluating" them over the reals so that the outputs lie entirely within $C$ whilst agreeing the the above mentioned canonical mapping of rational functions.

A pure existence proof for this correspondence would be nice but a construction for a method of evaluation for any $q\in \overline{R(x)}$