Let $M$ be a holomorphically symplectic complex manifold, and $f: M \to X$ a holomorphic, birational contraction to a Stein variety $X$, contracting a subvariety $E$ to a point, and bijective outside of $E$.

When $M$ is quasiprojective, it is known that $E$ is Lagrangian with respect to the holomorphic symplectic form. This fact is proven as follows. First, we note that $E$ is isotropic with respect to the holomorphic symplectic form (this is one line argument). Then, using some deep results of MMP (due to Kawamata and Shokurov, I think), one shows that $E$ contains a rational curve. Then the dimension of $E$ is bounded from below, using the deformation theory of this curve.

I want to have this result without $M$ being quasiprojective. I can show that $E$ is isotropic, and it is Lagrangian if and only if $E$ contains a rational curve. However, I don't see how to prove existence of a rational curve: the MMP arguments are way too complicated, and don't work in this context. I would be very grateful for any pointers which could be used to show existence of a rational curve on the exceptional set.

In algebraic context, the existence of rational curves was discussed there: Does a resolution of a rational singularity have rationally connected fibers? and Jason Starr replied that the fiber is rationally connected, "as follows from Elkik together with Hacon-McKernan's proof of Shokurov's conjecture". This is hard, but existence of a rational curve is weaker, and I hope it can be proven without that much effort.