Conditional entropy: $H(X | Y)$ large implies $H(X | Y, X \neq Y)$ large?
This might be a partial answer to your question:

Let $X$ and $Y$ be random variables with the same range. Let $Z$ be the indicator of the event $X \ne Y$. By the chain rule,
$$H(X|Y,Z) = H(X,Z|Y)-H(Z|Y).$$
Since $Z$ is determined by $X$ and $Y$, we have $H(X,Z|Y)=H(X|Y)$. Now, by the definition of conditional entropy, $$H(X|Y,Z) = \underset{z \leftarrow Z}{\mathbb{E}}[H(X|Y,Z=z)] = \mathbb{P}[X \ne Y] \cdot H(X|Y, X \ne Y) + \mathbb{P}[X = Y] \cdot H(X|Y, X = Y).$$ Clearly $H(X|Y, X = Y)=0$. Combining the above gives $$\mathbb{P}[X \ne Y] \cdot H(X|Y, X \ne Y) = H(X|Y) - H(Z|Y).$$
Now we have an equation linking the quantities we want. We have $$0 \leq H(Z|Y) \leq H(Z) = H_2(p) \leq 1,$$ where $p = \mathbb{P}[X \ne Y]$. In particular, $$p \cdot H(X|Y,X \ne Y) \leq H(X|Y) \leq p \cdot H(X|Y,X \ne Y) + H_2(p).$$