I know that $Q_n = Q_{n-1} \times K_2$. So I let $U=\{u_1,u_2,\dots, u_{n-1}\}$ be the set of vertices of the first copy of $Q_{n-1}$ and $V=\{v_1,v_2,\dots, v_{n-1}\}$ be the set of vertices of the second copy of $Q_{n-1}$.
I think of proving this by induction
Base : $n=2$, $Q_2$ is a $C_4$ so it's certainly Hamiltonian
Inductive step: Assume $Q_k$ is Hamiltonian. Show $Q_{k+1}$ is Hamiltonian. I know that the Cartesian product of any 2 Hamiltonian graphs is another Hamiltonian. but $Q_{k+1}=Q_k \times K_2$ and $k_2$ isn't Hamiltonian.
Do a Hamiltonian traversal of one copy of $Q_k$ except for the last edge (i.e. a Hamiltonian path, not cycle, through the first cube), then cross over to the second copy of $Q_k$ and traverse it in reverse order from the first copy, and then close the big cycle.