It now seems rather clear (but subject to more detailed assessment) that the answer to the first part ("can/should one,...") is yes.
As to the "how" part of the question, I've made use of the very elegant algorithm presented by Martin Roberts
in his Mathematica.stackexchange answer https://mathematica.stackexchange.com/questions/143457/how-can-one-generate-an-open-ended-sequence-of-low-discrepancy-points-in-3d
now using $d = 36$ and $d = 64$ (setting, as suggested, the supplementary parameter $\alpha_0=\frac{1}{2}$) in the two cases ("two-re[al]bits" and "two-qu[antume"]bits) of particular interest. I seem to be getting much stronger convergence than when I simply use (unstructured) random normal variates.
To convert from the points ($p$) the Roberts algorithm yields, that are uniformly distributed over $[0,1]^d$, to the unit normal variates required in the random matrix generation, I employ the Mathematica command
InverseCDF[NormalDistribution[0, 1],p].
The Mathematica code I am employing in the $d=36$ (two-rebit) case is
sp2 = x /. Solve[x^(37) == x + 1, x][[1]];
G = Array[1, 36];
Do[G[[i]] = 1/2 + i95/sp2^i, {i, 1, 36}];
rB = 0;
Do[hw = TimeUsed[];
P = InverseCDF[NormalDistribution[0, 1], N[FractionalPart[G]]];
Y1 = (Orthogonalize[ArrayReshape[Take[P, {1, 16}], {4, 4}]] +
IdentityMatrix[4]).ArrayReshape[Take[P, {17, 36}], {4, 5}];
z = Partition[Y1.Transpose[Y1], {2, 2}];
If[PositiveDefiniteMatrixQ[
ArrayFlatten@{{z[[1, 1]], z[[2, 1]]}, {z[[1, 2]], z[[2, 2]]}}] ==
True, rB = rB + 1];
If[Mod[i95, 500000] == 0,
Print[{TimeUsed[] - hw, i95, rB, rB/i95, N[rB/i95, 20]}];
{i95,rB} >> LowDiscrepancyBuresTwoRebitsSave], {i95, 1, 1000000000}]
The ratio of the variable rb to the variable i95 is the quantity of interest (that is, the separability probability with respect to Bures measure of the two-rebit states).