The answer is yes if "w(r) has exactly the same asymptotic decay as g(r)" means that $|w(r)|=O(e^{-kr}/\sqrt r)$, and then, instead of the condition "that $w(r)$ decays exponentially fast" it would suffice if we just assume, a priori, that (say) $w(\infty-)=0$ (and $k>0$). Everywhere here, $r\to\infty$.

Indeed, letting $c(r):=w'(r)r$, we rewrite as the ODE as $$c'(r)-k^2c(r)=-h(r):=-rg(r)=O(e^{-kr}\sqrt r),$$ whence, for some real constant $c_1$, \begin{equation} c(r)=c_1 e^{k^2r}+e^{k^2r}\int_r^\infty e^{-k^2s}h(s)\,ds =c_1 e^{k^2r}+O(e^{-kr}\sqrt r), \end{equation} by the l'Hospital rule, say. The condition $w(\infty-)=0$ implies that $w'(r_k)\to0$ and hence $c(r_k)=w'(r_k)r_k=o(r_k)$ for some sequence $(r_k)$ such that $r_k\to\infty$ as $k\to\infty$. So, $c_1=0$, $c(r)=O(e^{-kr}\sqrt r)$, $w'(r)=c(r)/r=O(e^{-kr}/\sqrt r)$, and hence \begin{equation} w(r)=-\int_r^\infty w'(s)\,ds =O\Big(\int_r^\infty \frac{e^{-ks}}{\sqrt s}\,ds \Big) =O(e^{-kr}/\sqrt r), \end{equation} again by the l'Hospital rule. $\Box$

This answer provides a slight modification of the excellent answer by Carlo Beenakker, to get $g>0$ on the entire interval $(0,\infty)$: if \begin{equation} w(r)=e^{-k r} \left(32 k \sqrt{r}-\frac{2}{\sqrt{r}}\right) \end{equation} and \begin{equation} g(r)=\frac{e^{-k r} \left(128 k^2 r^2-16 k r+1\right)}{2 r^{5/2}} \end{equation} for some $k\ne0$ and all real $r>0$, then $g>0$ on $(0,\infty)$, the equation $w''(r)+\frac1r\,w'(r)-k^2w(r)=-g(r)$ is satisfied for all real $r>0$, and $g(r) = O(e^{-kr}/\sqrt r)$ as $r\to\infty$, whereas $ w(r)\sim 32k e^{-k r} \sqrt r\ne O(e^{-kr}/\sqrt r)$ as $r\to\infty$.