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

The nice answer by user michael still needs some brushing up, which is what I have done here. I think this could help readers appreciate michael's answer, the central point of which is using the Vitali covering lemma:

Let $A\subseteq\R^d$ be such that $\mu(\R^d\setminus A) = 0$ and $\lambda(A) = 0$. For any real $\epsilon>0$, there is an open set $O_\ep$ such that $ A \subset O_\ep$ and $\lambda(O_\ep) < \epsilon$. Take any real $c>0$ and let
\begin{equation}
A_c:=\{x\in A\colon \liminf_{r\downarrow0}\frac{\mu(B(x,r))}{\lambda(B(x,r))}< c\}.
\end{equation}
For each $x\in A_c$ pick $r_x>0$ such that
\begin{equation}
B(x, r_x) \subset O_\ep\quad\text{and}\quad \frac {\mu(B(x,5r_x))}{\lambda(B(x,5r_x))}<c.
\end{equation}
By the Vitali covering lemma, there is a countable set $J\subseteq A_c$ such that the balls $B(x,r_x)$ are disjoint for distinct $x\in J$ and
\begin{equation}
A_c \subseteq \bigcup_{x\in J} B(x,5r_x).
\end{equation}
But then
\begin{multline}
\mu(A_c) \le \sum_{x\in J}\mu(B(x,5r_x))\le c \sum_{x\in J} \lambda(B(x,5r_x)) \\
\le c\,5^d \sum_{x\in J} \lambda(B(x,r_x))
\le c\,5^d \la(O_\ep)\le c\,5^d\ep.
\end{multline}
Letting $\ep\downarrow0$, we see that $\mu(A_c)=0$, for all real $c>0$. This implies that $\mu$-a.e.
$$\frac{\mu(B(x,r))}{\lambda(B(x,r))}\to\infty \quad \mbox{ as } \ r\downarrow0.$$