The answer is no. In fact, something far more dramatic can happen: there exist Radon measures $\mu$ on $\mathbb{R}^d$ such that for $\mu$ almost all $x$, for any arbitrary Radon measure $\nu$, the normalized restriction of $\mu$ to $B(x,r)$ "looks like" $\nu$ (it is arbitrarily close in the weak$^*$ topology).
To be more precise, given a measure $\mu$, a point $x$ in its topological support, and $r>0$, let
$$
\mu_{x,r}(A) = \mu(rA+x)/\mu(B(x,r)).
$$
Note that $\mu_{x,r}$ is obtained by magnifying $\mu$ around $x$ by a factor $r^{-1}$ and normalizing so that the unit ball gets unit mass. Weak$^*$ accumulation points of $\mu_{x,r}$ as $r\downarrow 0$ are called tangent measures of $\mu$ at $x$.
Now O'Neil [O'Neil, Toby. A measure with a large set of tangent measures. Proc. Amer. Math. Soc. 123 (1995), no. 7, 2217--2220] constructed a Radon measure $\mu$ which has all Radon measures as tangent measures at almost all points. See also [Sahlsten, Tuomas. Tangent measures of typical measures. Real Anal. Exchange 40 (2014/15), no. 1, 53--79].
In particular, let $\nu_\delta$ be a smooth measure such that $\nu_\delta(B(0,2/3)) \le \delta \nu_\delta(B(0,1))$. Since $\mu$ has $\nu_\delta$ as a tangent measure at almost all points, this means that for $\mu$-almost all $x$ there are arbitrarily small $r$ such that $\mu(B(x,r/2)) \le 2\delta \mu(B(x,r))$. As $\delta>0$ is arbitrary, $\mu$ fails to be locally doubling at almost all points, even if weaken the notion of locally doubling by requiring the doubling only for sufficiently small $r$ depending on $x$.