The statement appears to me to be false. The difficulty, from some point of view, is that in the spectral sequence going from $H^{-i}(G;\pi_j(M))$ to $\pi_{i+j}(M^{hG})$ an infinite number of torsion groups can conspire to make something rationally non-trivial.

As I understand it, the statement you are asking about says that if a finite group $G$ acts on a space $M$, and if both $M$ and the homotopy fixed point space $M^{hG}$ are simply connected, and if $M\to M_{\mathbb Q}$ is a rationalization of $M$ that is also a $G$-map for some action of $G$ on $M_{\mathbb Q}$, then the induced map $M^{hG}\to (M_{\mathbb Q})^{hG}$ is also a rationalization.

Let's make a counterexample in which $G$ acts trivially on $M$ (and on $M_{\mathbb Q}$). In this case $M^{hG}$ is the function space $Map(BG,M)$. I claim that it is enough if we can find a simply connected space $M$ and a finite group $G$ such that the based function space $Map_\ast(BG,M)$ is simply connected but not rationally trivial. If so, then the fibration sequence
$$
Map_\ast(BG,M)\to Map(BG,M)\to M
$$
shows that $Map(BG,M)$ is simply connected, and also in the fibration sequence
$$
Map_\ast(BG,M)_{\mathbb Q}\to Map(BG,M)_{\mathbb Q}\to M_{\mathbb Q}
$$
the right hand map is not a weak equivalence. On the other hand, in the fibration sequence
$$
Map_\ast(BG,M_{\mathbb Q})\to Map(BG,M_{\mathbb Q})\to M_{\mathbb Q}
$$
the right hand map is a weak equivalence because $H^j(BG;V)=0$ for $j>0$ and $V$ a rational vector space. It follows that
$$
Map(BG,M)_{\mathbb Q}\to Map(BG,M_{\mathbb Q})
$$
is not a weak equivalence.

To come up with such an $M$ and $G$ we can use the Atiyah-Segal completion theorem. The basic idea is to take $M$ to be $BU$, but I have to modify this a little to make $Map_\ast (BG,M)$ simply connected.

Start with $BU$, whose homotopy groups are $\pi_{2k}\cong \mathbb Z$ with complex conjugation acting by $+1$ when $k$ is even and by $-1$ when $k$ is odd. Localize it by inverting $2$, and split the result as a product of two factors according to that conjugation action. $M$ will be the factor corresponding to $-1$, so its homotopy groups are $\pi_j\cong \mathbb Z[1/2]$ if $j$ congruent to $2$ mod $4$.

Let $G$ be the dihedral group of order $6$. Then $H^i(BG;\mathbb Z[1/2])$ is trivial when $i$ is not a multiple of $4$ and is of order $3$ if $i>0$ is a multiple of $4$. It follows that $Map_\ast(BG,M)$ is simply connected and that $\pi_2Map_\ast(BG,M)$ is the inverse limit of larger and larger finite $3$-groups. By Atiyah-Segal (which describes the homotopy groups of the related space $Map_\ast(BG,BU)$) there is actually a copy of the $3$-adic integers here, so the group is rationally nontrivial.