Your question is more about ordinary algebra than homological algebra: you don't really need the chain complex structure on $C$ at all, except to induce a cochain complex structure on $C^*$. So henceforth I'll pretend $C$ is just an $R$-module.
Fact. If $C = R^{\oplus \kappa}$ is the free left $R$-module on $\kappa$-many generators and $G$ is a left $R$-module, then $\textrm{Hom}_R(C, G) \cong G^{\times \kappa}$ as left $R$-modules, naturally in $G$.
Fact. The product of finitely many left $R$-modules is also the direct sum, i.e. $A \times B \cong A \oplus B$ naturally in $A$ and $B$.
Proposition. If $\textrm{Hom}_R(C, G)$ is a free left $R$-module for all free left $R$-modules $C$, then $G$ must be a free left $R$-module.
Proof. Take $C = R$, then $\textrm{Hom}_R(C, G) \cong G$, so $G$ is a free left $R$-module.
Proposition. If $C$ is a finitely generated free left $R$-module, then $\textrm{Hom}_R(C, G)$ is a free left $R$-module for all free left $R$-modules $G$ (not necessarily finitely-generated!).
Proof. If $C = R^{\oplus n}$, then $\textrm{Hom}_R(C, G) \cong G^{\times n} \cong G^{\oplus n}$, and the direct sum of free left $R$-modules is a free left $R$-module.
There's not much more we can say in general. For example, if $R$ is a field, then all $R$-modules are free. So there's nothing to prove. On the other hand, for $R = \mathbb{Z}$, $C = \mathbb{Z}^{\oplus \aleph_0}$, $G = \mathbb{Z}$, we get $\textrm{Hom}_\mathbb{Z} (\mathbb{Z}^{\oplus \aleph_0}, \mathbb{Z}) \cong \mathbb{Z}^{\times \aleph_0}$, but the latter $\mathbb{Z}$-module is not free.