The space of genus 0 $n$-marked stable maps of degree r to $Q$ is proper, and so the image of the evaluation map $M_{0,n}(Q,r)\rightarrow Q^n$ is going to be a closed subset of $Q^n$. It remains to show that there is a rational normal curve through a generic set of $r+2$ points on $Q$; there should be a lot of ways to do this (I had in mind degenerating $Q$ to a double plane, where this statement can be confirmed by explicit calculation, and then arguing that a rational normal curve in this double plane deforms back to a rational normal curve in Q.)

If for every partition of your set of points into $S_1$ and $S_2$, the intersection of the span of $S_1$ and the span of $S_2$ (which will be a single point by the assumption of linear general position) does not lie in the quadric, then there will be a rational normal curve in $Q$ through your points. This is because there must be a genus 0 stable map of degree $r$ through your points, and this must necessarily be an embedding of a rational normal curve (all possible degenerate cases can be excluded by our conditions.) Unfortunately there are a lot of details to write out...

@Mark Indeed, I didn't see that...

You should get a counterexample from taking any projective variety and gluing two points.

Is there any reason to believe something like this should be true? I don't see why there should be any natural map between the two and it's false for $S$ empty...

I believe the Coolidge-Nagata conjecture is now known, see arxiv.org/abs/1502.07149

What's your action of W on the flag variety? I don't know of any natural ones (because I don't know of any natural embeddings of W into G.)

Relevant: mathoverflow.net/questions/56011/…

@poorna But that follows from Pic(X)=Z[L]; I think as long as that's true, it doesn't matter if C is generic in the linear system.

Doesn't Lazarsfeld actually prove that any smooth curve in $|L|$ is Brill-Noether general (Corollary 1.4 in his paper)?