Another comment on the local cohomology part of the question: Besides Karl's perfect answer one might also point out that the twisting does nothing to the cohomology group. The twisting line bundle is trivial in a neighbourhood of the point, so locally near the point $\mathcal F(n)\simeq \mathcal F$.

On the other hand, your question is exactly about local cohomology: If $Y$ is quasi-projective and its projective closure is $X$, then the groups $H^i(X,\mathcal F(n))$ and $H^i(Y,\mathcal F(n))$ are part of a long exact cohomology sequence with the third vertex being local cohomology supported on $X\setminus Y$.

Finally, ploughshare's example is actually just an explicit manifestation of Karl's comment via the (equivalent of the) above sequence: $\mathbb A^2$ is affine, so the cohomology of $\mathbb A^2\setminus {0}$ is the same as the local cohomology at the point $0$. There is a shift of indexes, so the non-vanishing of $H^1(\mathbb A^2\setminus 0, \mathcal O)$ is equivalent to the non-vanishing of $H^2_0(\mathbb A^2, \mathcal O_0)$ which follows from the theorem of Grothendieck Karl mentioned.