For $Q_0$, there is a HC. For $Q_1$, there's no HC. For $Q_2$, it forms $C_4$ so there's a HC. After this, I try to reason using Dirac's theorem of sufficiency for HC,
An n-vertex graph in which each vertex has degree at least $n/2$ must have a Hamiltonian cycle.
For the cube graph, each vertex has degree of $n$, and the number of vertices is $2^n$, so I realise I can't apply that. Neither can I apply Ore's theorem.
Is there a way to find the answer?