Doubt about the module of coinvariants

Question

Suppose that $G$ is group, $R$ is a ring (commutative, associative with $1\neq0$) and $M$ is a left $RG-$módulo such that $M$ is free as a left $R-$module. Is it true that the module of coinvariants $M_{G}$ is free? $$M_{G}:=M/Q,$$ where $Q$ is the submodule generated by the elements $gm-m$ for all $g\in G$ and $m\in M$.