Suppose $M$ and $M'$ are two $R$-moules (I am most interested in the case of $R$ a DVR). If $M\otimes M'$ is a semi-simple module (i.e., every submodule is a direct summand) then is it true that the tensor factors $M$ and $M'$ are semi-simple ?
I.e., if this is not true then it would be nice to see a counter-example.
No; if $R$ is a DVR with uniformizing paramenter $t$, the $R$-module $(R/tR) \otimes (R/t^2R) = R/tR$ is semisimple, but $R/t^2R$ is not.
I can't resist providing the following comment and references, even though it is likely not really relevant to the (already old by now) original post.
Quick background: Let $k$ be a field of char. $p>0$ and let $G$ be a(ny) group. Serre proved [Invent. Math. 116 (1994), no. 1-3, 513--530] that if $V$ and $W$ are finite dimensional, semisimple $kG$-modules and $$(\dim V - 1) + (\dim W -1) < p$$ then $V \otimes_k W$ is a semisimple $kG$-module. (The argument is quite nice - one reduces to alg. closed $k$, and replaces $G$ by a certain algebraic group over $k$ whose identity component is reductive. One then has to argue that $[G:G^0]$ has order prime to $p$, so one is reduced to consideration of connected reductive $G$. And that case is handled by some "weight combinatorics" and the linkage principle.)
In a subsequent paper [Semisimplicity and tensor products of group representations: converse theorems. With an appendix by Walter Feit. J. Algebra 194 (1997), no. 2, 496--520] Serre proved some "converse theorems". For example, he shows that $$V \otimes_k W \quad \text{semisimple} \implies V \ \text{semisimple if $\dim W \not \equiv 0 \pmod{p}$}$$ Examples (due to Feit, and included in the paper) show that one can't get rid of the assumption on $\dim W$.
Another example: Let $A,B$ be two non-semisimple modules over a ring whose tensor product is zero (e.g. $R = \mathbb{Z}, A = B = \mathbb{Q}/\mathbb{Z}$); this was discussed elsewhere.
