The answer is yes.
The proof is as follows: If there is a Hamiltonian cycle $f: \{0,...,2^{2^n}-1\} \rightarrow \{0,1\}^{2^n}$ of the hypercube where the number of edges of a given direction in the Hamiltonian cycle are the same, then there is such a bijection $\varphi$, by setting $\varphi_m(x)=\sum_{k=1}^m f_k(x)$, $m=1,2, ... 2^n$ (i.e. taking the cumulative sum of $f(x)$). Such a Hamiltonian cycle will be called a balanced Hamiltonian cycle.
My construction of balanced Hamiltonian cycles is inspired by this paper, which provides balanced Hamiltonian cycles for $n=1,2,3$. The construction for $n=2$ will serve as the base case of mathematical induction.
Assume the existence of a balanced Hamiltonian cycle for $n$, and make it a directed cycle. Let $g$ be a function on $\{0,1\}^{2^n}$ where $g(x)$ is the next vertex of $x$.
Let $h(x)$ be an involution on $\{0,1\}^{2^n}$ such that there are exactly $2^{2^n}/2^n$ values where $x$ and $h(x)$ differs at the $k$th bit, $k \in \{1 ... 2^n \}$. (The existence of $h$ will be proven later.)
Then there is a balanced Hamiltonian cycle on the hypercube of dimension $2^{n+1}$.
Let's identify the $2^{n+1}$-dimensional hypercube by the Cartesian product of the $2^n$-dimensional hypercube with itself. Let $(a_0, b_0)$ be a vertex of the $2^{n+1}$-dimensional hypercube. The sequence $(a_k, b_k)$ is defined as follows: if $k$ is odd, then $(a_{k+1}, b_{k+1})=(a_k, g(b_k))$; if $k=0 \text{ or } 2
(\text{mod } 2^{2^n+1})$, then $(a_{k+1}, b_{k+1})=(g(a_k), b_k)$; otherwise $(a_{k+1}, b_{k+1})=(h(a_k), b_k)$
One can see the function $k\rightarrow (a_k, b_k)$ plays the role of $f$ for $n+1$. The point is to analyze a cycle of length $2^{2^n+1}$, conclude that $(a_{2^{2^n+1}m}, b_{2^{2^n+1}m})=(g(g(a_{2^{2^n+1}(m-1)})),b_{2^{2^n+1}(m-1)})$ and it follows that $(a_k, b_k)$ traces out a Hamiltonian cycle (i.e. you need $m=2^{2^n-1}$ for the point to return to $(a_0,b_0)$). It's easy to check the Hamiltonian cycle is balanced.
Now we only to prove the existence of $h$.
If $n=2$, then such $h$ exists:
0000 - 0001
1110 - 1111
0010 - 0110
0011 - 0111
0100 - 1100
0101 - 1101
1000 - 1010
1001 - 1101
Assume the existence of $h$ for $n$. Choose a subset $K$ of $\{0,1\}^{2^n}$ with size $2^{2^n-1}$ that is closed under $h$ and there are exactly $2^{2^n-1}/2^n$ elements $x$ in $K$ where $x$ and $h(x)$ differs at the $k$th bit ($k \in \{1 ... 2^n \}$).
Define a function $p$ on $\{0,1\}^{2^n} \times \{0,1\}^{2^n}$: $p(a,b)=(a,h(b))$ if $b \in K$, otherwise $p(a,b)=(h(a),b)$. Then $p$ plays the role of $h$ for $n+1$.
Best Answer
I believe these are exactly odd $k$.
Indeed, it can be easily seen that if $k$ is even, then a cycle starting at $0^n$ can visit only those vectors that have even number of $1$'s, and so it cannot be Hamiltonian.
On the other hand, there exist long-run Gray codes where any consecutive $\leq n-3\log_2 n$ bit changes happening at distinct positions. So, for odd $k\leq n-3\log_2 n$ traversing such a code with step $k$ produces a Hamiltonian cycle in $H(n,k)$.
Also, for an even $n$, inverting every second element of a cycle in $H(n,k)$ produces a cycle in $H(n,n-k)$.
It remains to address the case of odd $n$ and odd $k>n-3\log_2 n$.