Well, (Q X (R-Q)) U ((R-Q) X Q) is measure 0 in the plane, unlike (R-Q) X (R-Q), even though both sets are dense and have the same cardinality. So perhaps the plane would still be red.
Now, if you let C be a non-measurable dense subset of R, and coloured every point in (R-C)X(C U Q) with fuschia paint mixed with day-glo glitter…
If (and that is a big if) you could paint all the rational points blue and all the irrational points red, then the plane would be red since the probability of a randomly chosen point being rational is 0. So a random light ray would almost certainly hit an irrational point. But this whole exercise, unless it is there to illustrate measure theory is, no pun intended, pointless.
To answer another point raised, yes the irrationals are also dense. A set of points is dense if for every element x and every positive number r, some element of the set is within distance r of x. In other words, every element can be approximated arbitrarily well by elements of the set. The rationals are dense, the irrationals are dense, the integers are not dense.
I think the plane might be orange in the first case. I knew the set had measure 0, but I wasn’t sure if it was dense (though I strongly suspected as much).
As for your second case…