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.
Added (7/7/2010): Having thought a bit about the non-Abelian case, I thought I'd add another non-vanishing theorem.
Theorem. Let $G$ be a Lie group admitting a faithful unitary representation, with $\pi_1(G)\neq 0, \mathbb{Z}$. Then there exists $M$ with $H^1(M, \underline{G})\neq 0$.
Proof. Let $\rho: G\to U(n)$ be the given faithful unitary representation, and let $M=U(n)/G$. Then $U(n)$ is a $G$-bundle over $M$, and it is non-trivial as $\pi_1(U(n))=\mathbb{Z}$ wheareas $\pi_1(G)$ cannot be a factor of $\mathbb{Z}$ by assumption. That is, $U(n)\not\simeq G\times M$ as $\pi_1(U(n))\not\simeq \pi_1(G)\times \pi_1(M)$. $\square$
This holds for e.g. compact Lie groups with the appropriate fundamental group; it seems likely that this argument can be strengthened by e.g. considering higher homotopy groups or using other results on the existence of faithful representations.