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 H1(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 H1(M) to S1. If the the image of f is pm1, 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.
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