For Abelian $G$ (that is, the product of a torus with $\mathbb{R}^n$), an argument identical to macbeth's comment gives that $H^n(M, \underline{G})=0$ for all $M, n>0$ iff $G\simeq \mathbb{R}^n$).
Explicitly, in the case $G\simeq \mathbb{R}^n$ the sheaf in question is fine; otherwise, if $G\simeq \mathbb{R}^n\times (S^1)^k$ then it fits into an exact sequence $0\to \mathbb{Z}^k\to \mathbb{R}^{n+k}(M)\to \underline{G}\to 0$, giving the claim.
As Dona Arapura suggests, your question is not well-defined at the moment for non-Abelian $G$.