Yes, it's a simple application of ultrafilters. Suppose that for fixed $d$, every finitely generated subgroup $H$ of a group $G$ has a faithful representation $j_H$ in $G_H=\text{GL}_d(K_H)$ for some field $K_H$. Then $G$ has a faithful representation into $\text{GL}(K)$, where $K$ is an ultraproduct of the $K_H$. Argue as follows: consider the lattice $I$ of all finitely generated subgroup. Consider an ultrafilter $\omega$ on $I$ containing, for every f.g. subgroup $H$, the set of subgroups containing $H$.
Then map $G$ into the ultraproduct $\ast^\omega(G_H)$ as follows: first extend $j_H$ to $G$ by defining $j_H(g)$ to be equal to 1 if $g\notin H$ (note that $j_H$ is a homomorphism only in restriction to $H$). Map $G$ into the product $\prod_H G_H$ by mapping $g$ to $(j_H(g))_H$. This is certainly not a homomorphism, but the composite map into the ultraproduct $\ast^\omega(G_H)$ is an injective homomorphism.
Now the ultraproduct $\ast^\omega(G_H)$ is canonically isomorphic to the group $\text{GL}_d(\ast^\omega K_H)$.
Finally this applies to free groups, but to many other groups, e.g. locally free groups and subgroups of ultraproducts of free groups, aka fully residually free groups. Note that if your group $G$ has a certain infinite cardinality, you can end up with a field of the same cardinality by restricting to the field generated by matrix entries of the image of your representation.