Let $\hat{R}\to R$ be a homomorphism of commutative unital rings and let $\hat{M}$ be an $\hat{R}G$-module for a group $G$. Does the $R$-module isomorphism $$H^n(G,\hat{M}\otimes R)\cong H^n(G,\hat{M})\otimes R$$ hold, where $\hat{M}\otimes R$ is an $RG$-module by extension of scalars?
If true, this must be well known, but I could not find a proper reference.
This isn't even true when $n = 0$. Let me write your morphism of rings as $f : R \to S$ because otherwise I'll keep thinking that $\hat{R}$ denotes some kind of completion. When $n = 0$ you want to know whether
$$\text{Hom}_{R[G]}(R, M \otimes_R S) \cong \text{Hom}_{R[G]}(R, M) \otimes_R S.$$
In other words, as in this MO question, you are trying to commute a hom and a tensor product. Homs are a kind of limit and tensor products are a kind of colimit, so you should expect that this isn't true in general, but that it might be true with some flatness hypothesis on $S$ as an $R$-module (or some projectivity hypothesis on $R$ as an $R[G]$-module, but you probably don't want that).
Explicitly, let $R = G = M = \mathbb{Z}$ (note that I'm naming objects in three different categories here...), where a generator of $G = \mathbb{Z}$ acts on $M$ by multiplication by $-1$. Here the LHS is the kernel of multiplication by $2$ acting on $S$, while the RHS is the the kernel of multiplication by $2$ acting on $\mathbb{Z}$ (which is zero), tensored with $S$ (which is still zero). These won't agree if $S$ has characteristic $2$ (which reflects the fact that in this case $S$ is not flat as an $R$-module).
If $S$ is flat as an $R$-module, then I think this is true for $n = 0$ if $G$ is finitely generated, but I expect you'll run into additional problems when $n > 0$.
For the same question with group homology replacing cohomology, everything is fine when $n = 0$ because you're now trying to commute a tensor product and another tensor product. For $n > 0$ there are still complications if $R$ isn't flat and you'll get a Grothendieck spectral sequence.
I want to add one very useful fact to complement the previous answer: The answer to your question is yes when $\hat R$ is a field.
One quick way to see that is to view it as a special case of the Kunneth formula, where one of the groups is trivial: $$ H^n(G\times1;\hat M\otimes R)\simeq H^n(G;\hat M)\otimes H^n(1;R). $$ The tensor products are over $\hat R$; of course for the map to be an iso we are using that $\hat R$ is a field so tensor product is exact.
