I used to understand this stuff pretty well, but it's been a long time. I think the following answer is correct, but I'm not certain.
Since M is a rational homology sphere, the irreducible points of R are separated from the reducible points, so we can treat them separately.
[EDIT: This isn't true in general (consider the case where $M$ is a connect sum). $M$ being a QHS only guarantees that the "very reducible" points of $R$, with image in the center of $SU(2)$, are isolated from the irreducibles. So the argment below only works in a special case.]
I claim that (1) the irreducible part of R contributes zero to the homological intersection number of the Q's, and (2) the contribution of the reducible part of R is $H_1(M)$. I think claim (1) follows from
(a) the answer to your question 2, which is that the contribution of a submanifold of intersection is equal to the Euler characteristic of its normal bundle (normal to both the Q's);
(b) using symplectic structure to show that the normal bundle is isomorphic to the tangent bundle in this case; and
(c) the observation that SU(2) acts freely, so the Euler characteristic of an irred. component of R is zero.
For claim (2), the idea is to show that the intersection number is equal to the number of homomorphisms $f$ from the finite abelian group $H_1(M)$ to $S^1$. If the the image of $f$ is $\pm 1$, then it corresponds to a unique transverse point of $R$. Otherwise, both $f$ and $-f$ lie on a 2-sphere component of $R$. By an argument similar to the one above, this 2-sphere contributes its Euler characteristic, namely 2, to the homological intersection number.
I've skipped a few details, but I think the basic outline is correct.