What is known about the H-factor problem?
This can be considered as a type of read-twice constraint satisfaction problem. Unless I'm mistaken, it is quite easy to construct an "equality" gadget whenever $H$ has a double gap, if you allow constant constraints (requiring an edge to be in or out). Then you can use Schaeffer's Boolean CSP classification. So the problem then comes down to whether constants are important.

There is a big generalisation:

"Fanout limitations on constraint systems", Tomás Feder, Theoretical Computer Science, Volume 255, Issue 1-2, March 28, 2001, Pages 281 - 293.

http://www.sciencedirect.com/science/article/pii/S0304397599002881

Basically, for constraint languages $\Gamma$ (i.e. a finite set of Boolean relations) such that $\Gamma$ contains the constant relations $\{(0)\}$ and $\{(1)\}$, Feder considers the following problem: the input is a conjunction of constraints of the form $(v_{i_1},\dots,v_{i_k})\in R$ for some $R\in\Gamma$, such that each variable appears at most twice, and we ask if there is a satisfying assignment. Feder shows that if the constraint language contains a non-delta-matroid, the problem is equivalent to the unbounded-degree CSP. A set $H$ that has gaps (i.e. whenever Cornuejols' algorithm doesn't work) is a non-delta-matroid.

You can reduce read-once-or-twice to read-twice by doubling the instance and identifying the variables that were read exactly once.

Added later:

For your particular problem, I believe the assumption about constants is easy to get around. Let $\mathbb{N}=\{0,1,2,\dots\}$. For all $\mathcal{H}\subseteq 2^{\mathbb{N}}$ and all $d\in\mathbb{N}\cup\{\infty\}$, the following problem is either in P or NP-complete.


Name: Factor$(\mathcal{H},d)$.
Input: a simple graph $G$ of maximum degree at most $d$, and a set $H_v\in\mathcal{H}$ for each $v$.
Output: Is there a spanning subgraph of $G$ in which the degree of each vertex $v$ is in $H_v$?


Sketch proof:


If for all $H\in\mathcal{H}$, there are no gaps of length $\geq 2$ in $H\cap\{0,\dots,d\}$, the problem is tractable by Cornuejols' algorithm.
If for $H\in\mathcal{H}$ we have $0\in H$ or $H\cap\{0,\dots,d\}=\emptyset$, the problem is trivial: every graph has a 0-factor so all inputs are accepted.
Otherwise, since (2.) fails, there is a non-empty set $H\in\mathcal{H}$ with $0\notin H$ and $\min(H)\leq d$.
So if a vertex $v$ has degree $\min(H)$ and $H_v$, then all its edges must be in. So we can force edges to be in, and simulate constants,
and appeal to Feder's result.


This kind of gadgetry fails if we restrict the degrees of the input graph. For example let $H=\{1,4\}\subset\{0,1,2,3,4,5\}$,
and restrict the inputs to graphs that are regular of degree five. By symmetry, one cannot simulate constants.