I realize that this might not directly answer your question (about references), but I would like to outline a possible approach for showing NP-hardness without the 2-connected condition. There are two things that are missing: one is a proof of the NP-hardness of the 'source problem', so to speak, and the other is that I'm reducing to a 'colored' version of H-cut that may or may not be useful. As for the first bottleneck, I believe I have a proof in my mind that I am being lazy about formalizing, so I hope I will get around to that soon. I've thought some about reducing the colored version to the one you present, however, with little luck so far. I am also very curious about your proof in the event that H is 2-connected, could you possibly supply some details?
So the colored version is the following: each vertex in the graph is equipped with a list of colors from a palette P (a fixed, finite set). We are required to find a cut so that no partition induces a monochromatic copy of H, that is, there is no subset of |H| vertices that induces a copy of H, and the corresponding list of colors have a non-empty intersection.
Here's a reduction from a restricted variant of d-SAT, where d is |H|. (Notice that this obviously wouldn't work when d = 2).
The restricted variant of d-SAT is the following:
Every clause has either only positive or only negative literals, let me refer to such clauses as P-clauses and N-clauses respectively,
Every P-clause can be paired off with a N-clause such that the two clauses involve the same set of variables.
(I have some idea about why this seemingly restricted version might be hard - a very closely related restriction is hard, and I can imagine a reduction from there, although I could be easily mistaken!)
Given this problem, the reduction perhaps suggests itself. The graph has a vertex for every variable of the formula. For every clause C_i, induce a copy of H on the set of variables that participate in the clause, and add the color i to this set of vertices. This completes the construction.
Any assignment naturally corresponds to a cut:
L = set of all variables that were set to 0,
R = set of all variables that are set to 1.
The claim is that a satisfying assignment corresponds to a monochromatic-H-free cut.
In other words, (L,R), when given by a satisfying assignment, would be such that neither L or R induce a monochromatic copy of H. If L has such a copy, then notice that the corresponding P-clause must have had all its variables set to 0, which contradicts the fact that the assignment was satisfying. Conversely, if R has such a copy, then the corresponding N-clause must have had all its variables set to 1, contradiction again.
Conversely, consider any cut, and set the variables on one side to 1 and the other to 0 (notice that it doesn't matter which way you do it - given the kind of formula we're working with, an assignment and it's flipped version are equivalent as far as satisfiability goes). If a clause isn't satisfied by this assignment, then we can trace it back to a monochromatic copy of H on one of the sides, contradicting the monchromatic-H-freeness of the cut.
The reason one has to indulge in the coloring is because copies of H can interfere to create spurious copies of H that don't correspond to clauses, in a direct reduction attempt. Indeed, it fails - badly - even when H is something as simple as a path.
I've had no luck in getting rid of the colors, and I am not sure that I have made the problem any simpler. However, I do hope that - if correct - it might be a start.
