I think the answer is yes. Let $M$ be the Artinian free $R$-module.
Proof. Let $S$ be a basis of the free $R$-module M. On the contrary, if $S$ were infinite, then we could pick an infinite sequence of pairwise distinct elements from $S,$ e.g., $s_1, s_2,$ etc.
From this, we have the strictly decreasing sequence of $R$-submodules $M_1 \supsetneq M_2 \supsetneq \cdots,$ where $M_i = R \langle s_j \,|\, j \geq i \rangle.$ However, this contradicts the assumption that $M$ is Artinian.
Is the proof valid? If so, can we then conclude that Artinian vector spaces are finitely generated?