The answer is yes in the category of groups.  Suppose that $f: G \to H$ is a retraction with $G$ a free group.  Then there is a homomorphism $g: H \to G$ such that $fg = \mathrm{id}_H$.
Thus $g$ is injective and hence embeds $H$ isomorphically as a subgroup of $G$.  But any subgroup of a free group is free, so $H$ must be free.  The same proof works in the category of abelian groups also.