The answer is yes. We know G is reductive, take B=TU a Borel.
Decompose V into a direct sum of weight spaces for T: $V=\oplus V_\lambda$. I claim that C contains a non-zero wieght vector. Proof: Suffices to do this for T=Gm. If v=Σivi with t in Gm acting on vi by multiplication by ti, consider $\lim_{t\to \infty}t^{-N}(t\cdot v)$ (t-N is scalar multiplication, dot is action). Since C is closed under action, scalar multiplication and taking limits, this is a weight vector in C for appropriate N.
If v in Vλ and u in U, then uv-v is in $\oplus V_\mu$ where only wieghts μ greater than λ appear in the sum. Let v be a weight vector in C of maximal possible weight . Applying u to v and running the torus argument again, we see that for this maximalilty assumption to hold, v must be annihilated by all of U. Since stabaliser of v is reductive, containing U, it must contain derived group of G.
What above argument really shows is that if λ is any weight appearing in V, then there exists some weight $\mu\geq \lambda$ with μ the weight of a one dimensional rep of G. (here we use C spans V). This implies that the only weights appearing in V are those corresponding to one dimensional representations of V, so V is a direct sum of characters.