For every vertex $v$ of $Q_n$, let $v'$ be the antipodal (opposite) vertex, consider the unique path in $T$ from $v$ to $v'$ and orient its first edge away from $v$. Since $T$ has fewer edges than vertices, some edge, say $xy$, has been assigned two orientations. The (combinatorial) distance between two antipodes of $Q_n$ is $n$ in $Q_n$, so it is at least $n$ in $T$, and the unique path in $T$, $x'...yx...y'$, from $x'$ to $y'$ has length at least $2n-1$. Finally, since $xy$ is an edge in $Q_n$, so is $x'y'$; adjunction of the latter to the unique path in $T$ from $x'$ to $y'$ yields the required cycle.