The solution decays like $r^{2-n}$ if $n \geq 3$. Setting $u=r^{\frac{1-n}{2}} v$, then $$v''-\frac{(n-1)(n-3)}{4r^2} v +g(r)v=0$$ with $g=4q$. This gives the result if $n=3$ since $rg(r) \in L^1$ and then all solutions of the above equation behave like $1,r$ as $r \to \infty$ so that the decaying solution behaves like $r^{-1}$. If $n >3$ one obtains the result by using the Liouville transformation for the equation satified by $v$, as in Chapter 6 of the book by Olver, Asymptotics and special functions (with some care because the functions $\psi_{f,g}$ does not belong to $L^1$). I can write down the details if somebody is interested but I do not see a short way as above.
EDIT The case $n>3$ goes as follows. Let $f(r)=(n-1)(n-3)r^{-2}$ and $\phi(r)=\int_1^r \sqrt{f(s)}ds=\frac{\sqrt{(n-1)(n-3)}}{2}\log r$. The function $w(s)=f^{\frac 14}(\phi^{-1}(s)) v(\phi^{-1}(s))$ satisies the equation $$w''(s)=\left (1+\frac{\psi_{f,g}(\phi^{-1}(s))}{\sqrt{f(\phi^{-1}(s))}}\right )w(s)$$ with $$\psi_{f,g}(r)=f^{-\frac 14}(-\frac {d^2}{dr^2} +g)f^{-\frac 14}$$ (see the book of Olver for this way). Performimg all computations one obtains $$ w''(s)= \left ( \frac{(n-2)^2}{(n-1)(n-3)} +O(e^{-\gamma s}) \right )w(s)$$ for some $\gamma>0$ (since $g$ decays more than quadratically). Then the decaying solution $w$ behaves like $w(s) \approx e^{-\frac {n-2}{\sqrt{(n-1)(n-3}}s}=r^{1-\frac n2}$. Coming back first to $v$ and then to $u$ one gets $u \approx r^{2-n}$.