So in that case (of a single zero-measure box, all of whose sub-boxes lying in the sub-copula have measure zero), you can do it.
One way to do it is to take a copula extending the sub-copula and then adjust. The way that I would do this is consider each atomic box $D$ from the sub-copula that intersects $\partial B$ [ by atomic, I mean a box of the form $[a,b]\times [c,d]$ with $a,b$ consecutive elements of $\mathcal S_1$ and $c,d$ consecutive elements of $\mathcal S_2$ ]. I would then uniformly the distribute mass from $D$ to $D\setminus B$. This gives a new copula which agrees with the previous copula on all of the atomic boxes, hence is an extension of the copula.
This can also be made to work by induction if there are a number of boxes to be assigned zero measure with gaps between. But it's not hard to see it can fail if one of the atomic boxes lies in the union of the zero-measure boxes, but not in any single zero-measure box.
More generally, I think you need a version of Hall's "Marriage Lemma". This is sometimes stated in terms of "men" marrying "women" (or points on the $x$-axis being coupled with points on the $y$-axis). I will use that language here, as it is easy to read, even though it suffers from lacking societal awareness. If there are the same number of men and women (corresponding to there being a probability distribution on the $x$- and $y$-axes), then there is a matching (corresponding to a copula) if and only if each subset $A$ of the men is collectively friendly with (i.e. the pair lies outside the zero-measure boxes) at least $|A|$ women; and each subset $B$ of the women is collectively friendly with at least $|B|$ men.
I would have to think in a bit more detail to see if I could formulate the precise condition analogous to the marriage condition in terms of copulae. Also, the Hall Marriage Lemma deals with two axes. One way to think of this in terms of the Hall Marriage Lemma is in terms of bipartite graphs: there are edges between a man and a woman if they are compatible. One is looking for a bijection. There are hypergraph versions of the Hall Marriage Lemma, which might allow one to deal with $n$-dimensional copulae, but for the moment, I have no idea how this would work.
EDIT: More details added in response to Q2 added in OP
Yes. I agree with your interpretation of my answer. One way to understand a copula is as a probability measure $\mu$ on $[0,1]^2$ such that $\mu(A\times [0,1])=\mu_1(A)$ and $\mu([0,1]\times B)=\mu_2(B)$ for any subsets $A$ and $B$ of $[0,1]$ where $\mu_1$ and $\mu_2$ are the probability distributions on the first and second coordinates respectively. (This is also called a coupling of the two measures).
In this language a subcopula is a measure on $[0,1]^2$ which is only defined on a sub-algebra $\mathcal B_1\otimes \mathcal B_2$ where $\mathcal B_1$ is the $\sigma$-algebra generated by intervals with your endpoints $\mathcal S_1$ and $\mathcal B_2$ is the $\sigma$-algebra generated by intervals with intervals with your endpoints $\mathcal S_2$. Another way to say this (at least in the case of finite $\mathcal S_1$ and $\mathcal S_2$) is that this $\sigma$-algebra consists of all unions of atomic rectangles. A subcopula is then extended to a copula if there is a measure on $[0,1]^2$ that agrees with the subcopula measure on the $\sigma$-algebra $\mathcal B_1\otimes \mathcal B_2$.
If $\bar\mu$ is a copula extending the subcopula $\mu$; and if the union of the zero measure rectangles does not contain any atomic rectangle, let $Z$ denote the union of the zero measure rectangles. You can then define a new copula extension by
$$ \lambda(S)=\sum_{\mathcal A\text{ atomic}} \bar\mu(S\cap A)\frac{\text{Area}(A\cap S\cap Z^c)}{\text{Area}(A\cap Z^c)}. $$
The condition ensures that you are not dividing by zero. To see that $\lambda$ is a copula, it suffices to check that it agrees with $\bar\mu$ on the atomic rectangles. If $R$ is an atomic rectangle, the only contribution to the sum comes from $A=R$. and that term is $\bar\mu(R)\text{Area}(R\cap Z^c)/\text{Area}(R\cap Z^c)$, which is equal to $\bar\mu(R)$.
To see that no mass is put on $Z$, if you substitute $S=Z$, then the numerator of the fraction is zero, so that $\lambda(Z)=0$. This finishes the proof.
There is nothing about this proof that is two-dimensional, so as long as the union of your zero-measure boxes does not contain any atomic boxes, this argument is fine.