In your example you obtained two linearly independent solutions with different behavior: $\cot(1/r)\sim r,\; r\to\infty$, so your second solution is $O(r^{-2})$.
This is the general pattern if you assume that your perturbation is analytic at $\infty$. Write your equation as $$f'=f^2+q(r),\quad q(r)=r^{-4}\sum_{0}^\infty a_kr^{-k}.$$ When this is so, make the change of the variable $r=1/\zeta$, $y(\zeta)=w(1/\zeta)$ in your linear equation for $w$ ($f=-w'/w$) and obtain $$\zeta^2y''+2\zeta y'+\zeta^2(q_0+\ldots)y=0.$$ The indicial equation $\rho(\rho-1)+2\rho=0$ has roots $0,-1$, so solutions $y$ can be bounded or behave as $c/\zeta$ near $\zeta=0$. Returning to your original equation, this means that some solutions $f$ are like $-1/r$, while others are like $O(r^{-2})$.
Which is the case for a particular initial condition is impossible to decide because your restriction $O(r^{-4})$ tells nothing about $q$ on a long finite interval.
My argument shows that $O(r^{-4})$ can be relaxed to $O(r^{-3})$ with the same conclusion.
My assumption that $q$ is analytic at $\infty$ is of course too strong, and can be relaxed, depending on your needs, but I don't think it can be completely dropped.