For $n\geq 8$, $K_n$ with one edge removed is not such an induced subgraph. It's clearly not a subgraph of $H(n,1)$ since the latter is bipartite.

Suppose that graph is an induced subgraph of $H(n,2)$, say it's spanned on strings $x_1,\dots,x_n$ with $x_{n-1},x_n$ not connected. Observe we may assume $x_1$ is the zero sequence, so the strings hence can be identified with at-most-two-element subsets of $[n]$ with pairwise symmetric differences of size at most two. First note that at least one of the sets has two elements (otherwise the induced graph would be complete), say $x_2$. But then at most two sets have exactly one element, ones corresponding to elements of $x_2$, let's say those are $x_3,x_4$ (we should also check for the case when $x_{n-1}$ or $x_n$ is one of those sets, but that case is easier). A little case work shows that if we have any four or more two-element sets which are pairwise intersecting, then they necessarily have a common element. Applying this to $x_2,x_5,\dots,x_{n-2},x_{n-1}$ and $x_2,x_5,\dots,x_{n-2},x_n$ have a common element. It follows $x_{n-1},x_n$ have a common element too, so they are connected in the graph. We need $n\geq 8$ here for there to be enough vertices, but this can be easily improved.

I believe a similar reasoning can be used to show that for a fixed $k$ and large enough $n$, $K_n$ without an edge is not an induced subgraph of $H(n,k)$ -- I can show this for $k=3$ (I won't bother writing that down in detail since it's a lot of case work), but for general $k$ we need some general kind of general result that if we have a family of set with large pairwise intersections, then there is some "core" contained in all of them.