$\newcommand{\N}{\mathbb N}
\newcommand{\R}{\mathbb R}
\newcommand{\B}{\mathcal B}
\newcommand{\F}{\mathcal F}
\newcommand{\la}{\lambda}
\newcommand{\si}{\sigma}
\newcommand{\Si}{\Sigma}
\renewcommand{\c}{\circ}
\newcommand{\tr}{\operatorname{tr}}$

Yes, the desired result holds for any mutually singular Radon measures $\mu$ and $\nu$, say on $\R^d$. Indeed, let $f$ and $g$ be, respectively, the Radon--Nikodym densities of $\mu$ and $\nu$ with respect to the measure $\rho:=\mu+\nu$. Then $fg=0$ $\rho$-a.e. For each real $r>0$, the function 
$$x\mapsto f_r(x):=\frac{\mu(B(x,r))}{\la(B(x,r))}$$
is the convolution $f*k_r$ of $f$ with the kernel function $k_r$ given by the formula 
\begin{equation}
	k_r(x):=\frac1{r^d}\,k(x/r), \quad k:=\frac1{\la(B(0,1))}\,I_{B(0,1)},
\end{equation}
where $I_{B(0,1)}$ is the indicator function of $B(0,1)$. So, $f_r\to f$ and, similarly, $g_r\to g$ $\rho$-a.e., for the similarly defined $g_r$; the convergence everywhere here is as $r\to0$. So, $f_r/g_r\to f/g$ $\rho$-a.e. and hence $\mu$-a.e., and $f/g=\infty$ $\mu$-a.e. Thus, $f_r/g_r\to\infty$ $\mu$-a.e.
\begin{equation}
	\frac{\mu(B(x,r))}{\nu(B(x,r))}=\frac{f_r(x)}{g_r(x)}\to\infty
\end{equation}
for $\mu$-almost all $x$, as desired.