If $G$ is compact, the inclusion $H(M^G) \to H(M)^G$ is an isomorphism. The inverse map is defined as follows: Take a class $\omega$ in $H(M)^G$ and lift it to a closed form $\alpha \in \Omega(M)$. Put $\beta = \int_{g \in G} g^{\ast} \alpha$, where the integral is with respect to Haar measure normalized to have volume $1$. Clearly, $\beta \in \Omega(M)^G$. Our lift will map $\omega$ to $[\beta]$. 

We must check that $\beta$ is closed, is a de Rham representative of $\omega$, and that its class in $H(M^G)$ is independent of the choice of $\alpha$. Since all $g^{\ast} \alpha$ are closed, so is $\beta$. Since all $g^{\ast} \alpha$ are de Rham representatives of $\omega$, so is $\beta$. Finally, let $\alpha' = \alpha + d \eta$ be another lift of $\omega$. Then
$$\int_{g \in G} g^{\ast} \alpha' = \int_{g \in G} g^{\ast} \alpha + d \int_{g \in G} g^{\ast} \eta$$
and $\int_{g \in G} g^{\ast} \eta$ is in $\Omega(M)^G$.

When $G$ is not compact, both injectivity and surjectivity can fail. 

<b>Failure of injectivity</b>:
Consider $M = \mathbb{R}$ and $G = \mathbb{Z}$ acting by translations. The $1$-form $dx$ is closed and $G$-invariant on $\mathbb{R}$, but has no $G$-invariant integral. So it gives a nonzero class in $H^1(M^G)$, but of course $H^1(M)^G \subseteq H^1(M) = 0$.

<b>Failure of surjectivity</b> Let $M = S^1$. For $\theta \in \mathbb{R}$, define
$$\tilde{\phi}(\theta) = \theta + \tfrac{1}{2} \sin \theta.$$
Then $\tilde{\phi}: \mathbb{R} \to \mathbb{R}$ descends to a diffeomorphism $\phi: S^1 \to S^1$, with repelling and attracting fixed points at $\theta = 0$ and $\theta = \pi$ respectively. We let $G = \mathbb{Z}$, acting by $\phi$. 

The map $\phi$ acts trivially on $H^1(S^1)$. I claim that there is no nonzero smooth $\phi$-invariant $1$-form on $S^1$. Suppose for the sake of contradiction that $\omega$ is a $\phi$ invariant $1$-form.

Let $\omega = g(\theta) d \theta$ by $\phi$ invariant, and suppose for the sake of contradiction that $g(\theta_0) \neq 0$ for some $\theta_0 \not \in \pi \mathbb{Z}$. Then 
$$g(\phi^n(\theta_0)) = \prod_{k=0}^{n-1} (\phi')(\phi^k(\theta_0))^{-1} g(\theta_0).$$
As $k \to \infty$, $\phi^k(\theta_0) \to \pi$ and $\phi'(\phi^k(\theta_0))^{-1} \to 2$. So $g(\phi^n(\theta_0)) \to \infty$ as $n \to \infty$, contradicting that $g(\theta) d \theta$ is supposed to be a continuous $1$-form.

<b>Final remark</b> When the action of $G$ is free, what you are calling $\Omega(M)^G$ and $H(M^G)$ are $\Omega(M/G)$ and $H(M/G)$. I couldn't find an example of failure of surjectivity in this case.