Your space $Q=Q_2$ is the cofibre of some map $\alpha\colon S^6\to S^3$, so $\alpha$ lies in the group $\pi_6(S^3)$, which is isomorphic to $\mathbb{Z}/12$ (see https://en.wikipedia.org/wiki/Homotopy_groups_of_spheres and the references cited there).  We therefore have a cofibration sequence 
$$ S^{13} \xrightarrow{\Sigma^7\alpha} S^{10} \to \Sigma^7Q \to S^{14} \xrightarrow{\Sigma^8\alpha} S^{11}, $$
giving rise to an exact sequence
$$ \pi_{11}S^7 \xrightarrow{(\Sigma^7\alpha)^*} \pi_{14}S^7 \xrightarrow{}
    [\Sigma^7Q,S^7] \to \pi_{10}S^7 
\xrightarrow{(\Sigma^8\alpha)^*} \pi_{13}S^7
$$
Inserting the calculations of the above homotopy groups, we get an exact sequence 
$$ 0 \xrightarrow{} (\mathbb{Z}/120)\sigma' \xrightarrow{}
    [\Sigma^7Q,S^7] \to (\mathbb{Z}/24)\nu \xrightarrow{(\Sigma^8\alpha)^*} 
    (\mathbb{Z}/2)\nu^2
$$
At the prime $2$ we see from Lemma 5.4 of Toda's "Composition methods in homotopy groups of spheres" that $\pi_6(S^3)$ is generated by $\nu'$ with $\Sigma^2(\nu')=2\Sigma\nu$.  It follows that $\Sigma^8\alpha$ is divisible by $2$, which forces the above map $(\Sigma^8\alpha)^*$ to be zero.  Thus, the first three terms above form a short exact sequence.  This means that the $2$-primary part of $[\Sigma^7Q,S^7]$ is $\mathbb{Z}/8\oplus\mathbb{Z}/8$ or $\mathbb{Z}/16\oplus\mathbb{Z}/4$ or $\mathbb{Z}/32\oplus\mathbb{Z}/2$ or $\mathbb{Z}/64$, and the $3$-primary part is $\mathbb{Z}/3\oplus\mathbb{Z}/3$ or $\mathbb{Z}/9$, and the $5$-primary part is $\mathbb{Z}/5$.  It may be possible to use Steenrod operations or Adams operations to resolve the extension problem, but I have not attempted that.