The following is only a partial answer. The number $H(n,k)$ is not a $2$-adic integer "for most $n$". I will stick to the case $k=2$ for convenience. I claim that there is a sequence $(a_j)_{j \geq 0} \in \{ 0,1 \}^{\mathbb{N}}$, with $a_0 =1$, such that if $n = \sum_{j=0}^r b_j 2^{r-j}$ is the binary expansion of an integer $n$, with $b_0 = 1$, and if $(b_j)_{j=0}^r \neq (a_j)_{j=0}^r$, then $$ v_2(H(n,2)) = - 2r + s, $$ where $s$ is the smallest integer such that $b_s \neq a_s$. In particular, $H(n,2)$ is not a $2$-adic integer, unless $n$ has the form $n = n_r = \sum_{j=0}^r a_j 2^{r-j}$ for some $r$. As $n_r \in [2^r,2^{r+1}[$, these possible counterexamples are very sparse, and this what I meant by "for most $n$" above ($n_r$ is the only possible exception in $[2^r,2^{r+1}[$).
Construction of $(a_j)_j$ : one constructs by induction sequences $(a_j,n_j,x_j)_{j \geq 0}$ such that $x_j = 2^{j-1} H(n_j,2)$ is a $2$-adic integer. One sets $a_0 = n_0 =1, x_1 =0$, and the induction step is given by
- $a_{j+1} \in \{ 0,1 \}$ is such that $a_{j+1} \equiv a_j + x_j \pmod 2$.
- $n_{j+1} = 2n_j + a_{j+1}$
- $x_{j+1} = 2^{j} H(n_{j+1},2)$
For example $n_0=1$, $n_1 = 3$, $n_2 = 6$, $n_3=13$, $n_4=27$, $n_5=54$, $n_6=109$, $n_7=219$. Correspondingly, $(a_j)_j = 1,1,0,1,1,0,1,1,...$
I do not know if $H(n_j,2)$ is a $2$-adic integer for only finitely many $j$ (for $j \leq 7$ this is a $2$-adic integer only for $j=0,1$). For example, for $j=7$, one has $v_2(H(n_7,2)) = v_2(H(219,2)) = -3$, which is unexpectedly large.