Almost everything is correct except for the reasoning at step (12). It is not true that $\partial_{p_k} (a_\mathbf{p}^\dagger a_\mathbf{p}) = 0$. In case differentiating an operator with respect to momentum is unfamiliar, note that it is defined just like every other derivative,
$$\frac{\partial}{\partial p_k} \mathcal{O}(\mathbf{p}) \equiv \lim_{\epsilon \to 0} \frac{\mathcal{O}(\mathbf{p} + \epsilon \hat{\mathbf{k}}) - \mathcal{O}(\mathbf{p})}{\epsilon}.$$
The operator $\mathcal{O}(\mathbf{p}) = a_{\mathbf{p}}^\dagger a_{\mathbf{p}}$ clearly depends on what $\mathbf{p}$ is, because it represents the number of particles with momentum $\mathbf{p}$. So its derivative doesn't vanish.
Instead, note that the error term is
$$\int d\mathbf{p} \, p^j \frac{\partial}{\partial p_k} (a_{\mathbf{p}}^\dagger a_{\mathbf{p}}) = \int d\mathbf{p} \, \frac{\partial}{\partial p_k} (p^j a_{\mathbf{p}}^\dagger a_{\mathbf{p}}) - \delta^{jk} a_{\mathbf{p}}^\dagger a_{\mathbf{p}}.$$
The second term vanishes by the antisymmetry of $\epsilon^{ijk}$, while the first term vanishes because it's just the integral of a total derivative. This is not any different than saying that
$$\int_{-\infty}^\infty dx \, \frac{df}{dx} = 0$$
except that the letter $x$ has been replaced with $p$. You might protest that $p^j a_{\mathbf{p}}^\dagger a_{\mathbf{p}}$ does not vanish as $p \to \infty$, but as the integration bounds go to infinity, the boundary terms essentially count the number of increasingly high-energy particles, which goes to zero for any finite-energy configuration. (This is a bit mathematically cavalier, and this kind of reasoning could fail in other contexts, but it's good enough for the free scalar field.)
To complete the derivation very explicitly, after normal ordering we're left with
$$Q_i=-2 i\epsilon_{ijk}\int\frac{d^3p}{(2\pi)^3} p^j a^\dagger_\vec{p} \partial_{p_k} a_\vec{p} = -2 i\epsilon_{ijk}\int\frac{d^3p}{(2\pi)^3} a^\dagger_\vec{p} p^j \partial_{p_k} a_\vec{p}.$$
Since $\epsilon_{ijk}$ is antisymmetric, only the part of the integrand antisymmetric in $j$ and $k$ contributes, giving
$$Q_i= -i\epsilon_{ijk}\int\frac{d^3p}{(2\pi)^3} a^\dagger_\vec{p} p^j \partial_{p_k} - p^k \partial_{p_j} a_\vec{p}.$$
This is exactly what we wanted to show.
Edit: to provide even further detail we'll show how the unwanted first two terms in (10) vanish. The stated reasoning in the OP is incorrect; the terms do not just vanish by symmetry. The first term is proportional to
$$I = \int d\mathbf{p} \, p^j a_p \frac{\partial}{\partial p_k} a_{-p}.$$
By changing variables from $p \to -p$, we find that
$$I = \int d\mathbf{p} \, (-p^j) a_{-p} \frac{\partial}{\partial (-p)_k} a_p = \int d\mathbf{p} \, p^j a_{-p} \frac{\partial}{\partial p_k} a_p = \int d\mathbf{p} \, p^j \left(\frac{\partial}{\partial p_k} a_p\right) a_{-p}$$
since lowering operators commute. Therefore, adding these two expressions,
$$2I = \int d\mathbf{p} \, p^j \frac{\partial}{\partial p_k} (a_p a_{-p}).$$
As before, we integrate by parts. The term with $\partial p^j / \partial p_k = \delta^{jk}$ again vanishes by the antisymmetry of $\epsilon^{ijk}$. The remaining boundary term can be argued to have no contribution for finite-energy contributions, i.e. its matrix elements between any two such configurations vanishes. The same argument can be applied to the second term in (10).
The whole derivation of the Mermin-Ho relation required a choice of $\hat{e}_1$ and $\hat{e}_2$ so that $\hat{n} = \hat{e}_1 \times \hat{e}_2$ everywhere. The construction of $\Omega$ requires them to be differentiable.
However, from the hairy ball theorem we know that it is impossible for $\hat{e}_1$ and $\hat{e}_2$ to be smooth and differentiable. Thus, there are places, where $\Omega$ is singular – but right there, the Mermin-Ho relation simply does not apply because our initial assumption, of $\hat{e}_1$ and $\hat{e}_2$ being smooth, does not hold.
To define the Mermin-Ho relation, poke a hole everywhere on the unit sphere where singularities occur and define $\hat{e}_1$ and $\hat{e}_2$ everywhere else apart from these holes. This way, one has excluded the singularities.
Best Answer
Yes, that's fine. You could write out each component individually if you want to assure yourself.
A more-intuitive argument would be to prove that line integrals of gradients are path-independent, and therefore that the circulation of a gradient around any closed loop is zero. The curl is a limit of such a circulation, and so the curl must be zero.