Here are some things that you probably know, but just to say something ... :
Suppose that $R$ is a $k$-algebra for a field $k$, and that $M$ is finite dimensional over $k$. Then $M$ is simple as a right $R$-module if and only
if $Hom_k(M,k)$ (with the transpose $R$-action) is simple as a left $R$-module.
Then $M\otimes_k N$ is a $k$-vector space, and since it is the universal recipient
of a $R$-bilinear pairing from $M\times N$, it is non-zero if and only if there is a $R$-linear pairing $M \times N \to k$, if and only if there is a non-zero $R$-module homomorphism $N \to Hom_k(M,k)$. If $N$ is simple, such a non-zero homomorphism has to be an isomorphism, and so we see that $M\otimes N \neq 0$ for simple $M$ and $N$ if and only if $N = Hom_k(M,k)$.
Of course my assumptions imply that $M\otimes_R N = Hom_R(Hom_k(M,k),N),$
and so this reduces to the "Hom" case that is motivating your whole question.
But maybe it suggests a way of thinking which might work in a more general
setting (trying to replace $k$ by some kind of "minimal" quotient of
$M\otimes_R N$). I didn't actually succeed yet in saying anything more general, though ... .