Yes, full CI is a much better way to determine what orbitals are likely to be significant than a Hartree-Fock computation. My usual approach is to do a full valence CAS and see what matters.
To put your numbers in perspective, if you compute full CI on water with 6-31G, your leading determinant has a coefficient of .978 and a weight (coefficient squared) of 0.956. The next most significant determinant has a coefficient of 0.053 and a weight of 0.003. Since water is most definitely a single reference problem, we can very safely say that you could have stopped including determinants long before you got to coefficients of 0.024409. Unless you want this MCSCF computation to specifically monitor the weight of determinants or have some very specific reason not to do so, cut down your active space to [2, 0, 0, 0, 0, 1, 1, 1] (weight >2%) or [2, 0, 0, 0, 0, 1, 2, 2] (weight > 1%). Having fewer determinants will make convergence much easier. If you want to do some correlated computation after this with Forte, definitely pare down the active space.
As for the missing B1g/B2g/B3g orbitals, you were right to exclude them. I was hasty and didn’t check the diagrams and symmetry arguments. Apologies if you have no clue what I’m talking about, but here’s “why” those orbitals don’t appear:
It looks like you want the ground state, which is singlet sigma plus gerade. If we assume the 1s bonding orbtial is occupied (1AgX) in all determinants, that doesn’t give us many ways to have any B_g orbitals occupied. We could have a A, a B1, a B2, and a B3, to maintain our sigma plus, but you can’t make gerade work. You could maybe have two electrons in the same B2 or B3, but then you either have both electrons in one spatial orbital in a spatially degenerate pair, or you fill both of those, which gives a negative bond order. Not favorable. You could send one electron to 1B2 and the other to 2B2, but excited antibonding orbitals? Also not likely. So on hindsight, it makes sense that B1g/B2g/B3g aren’t going to be important orbitals.