Let $G$ be a finite group, and $P$ a fg projective $\mathbb{Z}[G]$-module. By a theorem of Swan, $P\otimes\mathbb{Q}\cong\mathbb{Q}[G]^n$ for some $n$. Is the same true for $\mathbb{R}$? Clearly, $(P\otimes\mathbb{R})^n$ is stably free (hence free) for some $n$ since $K_0(\mathbb{Z}[G])$ is finite (again by Swan), but I don't think this implies anything I want.
If $P\otimes\mathbb{R}$ isn't free, then can we say anything about it? Does it contain a factor of $\mathbb{R}[G]$, for example? I vaguely recall the rank of $P$ has to be some multiple of $|G|$ (once more by Swan) but can't remember where I read this.
[Comment reposted as an answer]:
We have $P \otimes \mathbb{R} = (P \otimes \mathbb{Q}) \otimes_{\mathbb{Q}} \mathbb{R}$.
Since $P \otimes \mathbb{Q} \cong \mathbb{Q}[G]^n$ for some $n$, it follows that $P \otimes \mathbb{R} \cong \mathbb{R}[G]^n$.