Let $n \in \mathbb{N}$, $p$ a prime number, and $G$ a finite group of order coprime to $p$. Let $R = \mathbb{Z} /p^n \mathbb{Z}$ be the ring of integers mod $p^n$. Must $R[G]$ be semisimple?
As noted in the comments, since $R$ itself is not semisimple this can not be true.
What if $R$ is semisimple (and has characteristic $p$)? Is there some generalization of Maschke's theorem valid here?
In any case, whether the associative $R$ is commutative or not ( as long as $R$ has $1$) if every finitely generated unital $RG$ module is to to be completely reducible, then $|G|1_{R}$ has to be invertible in $R$. The proof I give may seem laboured, but I don't want to make unjustified assumptions.
For let $A$ be the kernel of the trivial representation of $RG$, that is, the representation of $G$ acting from the right on $R$ and with $rg = r$ for all $r \in R,g \in G$ (and with right regular action of $R$ on itself) Then $A = \{ x \in G: x = \sum_{g \in G} r_{g}g : r_{g} \in R, \sum_{g \in G}r_{g} = 0 \}$, the augmentation ideal of $RG,$ and a (proper) two-sided ideal. Suppose that $RG = I \oplus A$ where $I$ is a right ideal of $RG.$ Write $1_{R} = e +a $ with $e \in I$ and $a \in A$, a unique expression. As usual, $e^{2} = e \neq 0.$ Now $G$ acts trivially (from the right) on $(RG)/A,$ as can be verified directly anyway, so that for each $g \in G,$ we have $eg = e +b_{g}$ for some $b_{g} \in A.$ But $eg \in I,$ so we must have $b_{g} = 0$ for each $g$ and $eg = e$ for each $g \in G.$ It follows that $e = t \sum_{g \in G}g$ for some $t \in R.$ However, $e^{2} = e$ also gives $|G| t^{2} \sum_{g \in G}g = t \sum_{g \in G}g,$ so that $|G|t^{2} = t$ on comparison of the coefficient of $1_{G}$. However, there is no non-zero $r \in R$ such that $tr = rt =0,$ for $e$ acts as the identity on the trivial (right) $RG$-module $R,$ whereas if $rt =0$ in $R,$ we see that $e$ annihilates $r$ in the right action of $RG$ on $R$. Hence we have $|G|t = 1_{R}$ and $|G|1_{R}$ is invertible in $R$.
It also seems to me that if finitely generated $R$-modules are completely reducible, and $|G|$ is invertible in $R,$ then Maschke's Theorem should go through for finitely generated $RG$-modules, whether or not $R$ is commutative. For if we have such an $RG$-module $M,$ and an $RG$-submodule $U,$ then we may choose a complement $V$ to $M$ as $R$-module. We take an $R$-module homomorphism $\phi:M \to M$ which is projection onto $U$ with kernel $V$ Then for $t$ as above we take $ \theta = t \sum_{g \in G} g^{-1}\phi g$ which is now an $RG$-module homomorphism, still with image $U,$ whose kernel meets $U$ trivially, so is an $RG$-module complement to $U$ . We don't need $R$ to be commutative, but we only seem to need that elements of $R$ commute with elements of $G$, which is built into the definition of the group ring $RG$.
Maschke's Theorem does generalise to arbitrary (commutative) coefficient rings. If $|G|$ is a unit in $R$, then every $R[G]$-module is a direct summand of a module induced from the trivial subgroup. When $R$ is semisimple, this property ends up being equivalent to being projective.
I'm not sure about the non-commutative case.
You can find a version of Maschke's Theorem for group rings over arbitrary coefficient rings with 1, commutative or not, in Milies & Sehgal's An Introduction to Group Rings. See Theorem 3.4.7. It states that $R[G]$ is semisimple if, and only if:
(1) $R$ is a semisimple ring;
(2) $G$ is a finite group;
(3) $|G|$ is a unit in $R$.
