Sasha's statement is true for any pair of modules.

The center of $R$ is a commutative ring $S$. Since the endomorphisms of a simple module are a division algebra, whose center is a field, the action of $S$ on every simple module factors through some field, so the action of $R$ of course factors through an algebra over that field.

The kenel of a map to a field is a prime ideal $p$, and the map to the field factors through the residue field $k_p$.

So if we have two finitely-generated modules $M$ and $N$, their annihilators in $S$ are two prime ideals of $S$, $p_1,p_2$. If the ideals are distinct, then $S$ annihilates $M \otimes_R N$ since the action of $S$ factors through $k_{p_1} \otimes_S k_{p_2}=0$. The tensor product is zero because one ideal necessarily contains an element $e$ not in the other. In the residue field that element, since it's not in the ideal, has an inverse. Then $1= 1\otimes 1= e^{-1}e\otimes 1=e^{-1}\otimes e=e^{-1} \otimes 0 =0$.

If they are the same ideal, set $R'= R\otimes_S k_p$. It is now an algebra over a field. Apply Sasha's statement.