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. $\Box$