The answer is no. Take $G=\mathbb{Z}/2$ and $V=\mathbb{Z}/3$ with the nontrivial action of $G$. Then for even $n$ the module $Sym^n(V)$ is trivial, so $H^0(G,Sym^n(V))=\mathbb{Z}/3\neq 0$. For odd $n$ we have $Sym^n(V)=V$ and every $H^i(G,Sym^n(V))=0$.
One way to see this is this. Take $H=\mathbb{Z}$ and a surjective group map $f:H\to G$. Then $V$ becomes an $H$-module. Moreover, $V$ is trivial as a $\ker f$-module and $G$ acts as multiplication by $-1$ on $H^*(\ker f,V)$. So both rows of the Serre-Hochschild spectral sequence of $f$ will be identical, and if one is non-zero, so would be the other. On the other hand we have $H^*(H,V)=0$ and so both rows of the spectral sequence, which are just the cohomology $H^*(G,V)$, are zero.
(This is just the simplest possible example, but there are many more. In general there is no way to reduce the cohomology of tensor products/symmetric/alternating powers of a sheaf to the cohomology of the sheaf itself.)
[upd: slightly modifying the above construction one can get an example of a free module $G$-module $V$ with $G$ cyclic such that $H^*(G,V)=0$ and $H^0(G,Sym^2(V))\neq 0$.
Namely, take $V$ to be the $\mathbb{Z}$-module $\mathbb{Z}^2$ on which 1 acts as $A=\begin{pmatrix}0&1\\-1&1\end{pmatrix}$. Notice that the order of $A$ is 6, so we $V$ is a $G=\mathbb{Z}/6$-module. Notice also that $I-A\in SL_2(\mathbb{Z})$, so $H^*(\mathbb{Z},V)=0$.
Now consider the Serre-Hochschild spectral sequence as above. Both the rows of it contain $H^*(G,V)$ and so both must be 0 since $H^*(\mathbb{Z},V)=0$. On the other hand it is not too difficult to construct an invariant in $Sym^2(V)$: pass to the dual and then construct an invariant quadratic polynomial.
Using Serre-Hochschild for this may be an overkill. It is possible that one can do all this using just the standard cochain complexes.]
