Yeah this is one of the many reasons I've found the statement strange. I'm guessing they really meant that the symplectic form is non-degenerate and holomorphic. In this case, $\mathbb P^n$ certainly doesn't contradict this.

It follows from Grothendieck's vanishing theorem. that Ext group is the same thing as $H^{\dim X-i}(L^{-N}\otimes E(K_X))^*$, and this sheaf has support of dimension $\dim supp(E)$. If $\dim X-i>\dim supp(E)$, then this must vanish by Grothendieck's vanishing theorem.

But how could you determine the discriminant without knowing the lattices ahead of time? In the specific situation I'm in, I have quadratic forms on some finite groups, and I need to know which p-adic lattice corresponds to my specific p-group and its quadratic form. Is the answer really just that I can't determine this?

I acknowledged, after Jack's comments above, that your proof works for the usual definition of rationally connected, namely that two general points can be connected by a rational curve. However, I edited the question to more clearly reflect what I was asking which was why ANY two points can be connected by a chain of rational curves.

I was also wondering if there were other examples that come up in practice. My method above can be very quickly used to give a proof of derived equivalence of classical flops since the normal bundle of $\mathbb P^n$ is then $\mathcal O_{\mathbb P^n}(-1)^{n+1}$ and the analogous vanishing as above is clear. Are there other examples?

That's what I thought from trying to work this out, but do you know of any important examples and proofs of why this works for them? With the above example, I can show that $\mathcal O_Z/I^n$ has a unique module structure for $n\leq 3$, where $I$ is the ideal sheaf of $\mathbb P^n$ in $Z$. So up to the second-order infinitesimal neighborhood both completions are the same. But this is a brute-force method and to continue it I'd need to know that Ext$^1(Sym^k(T_{\mathbb P^n}),Sym^l(T_{\mathbb P^n}))=0$ for $k>l$ or $l>k$ (I forget which) which I don't know is true at the moment.