I think in order to prove the sharp result you need to use a covering theorem. The best option seems to be Vitali's covering theorem for Radon measures, which says the following: let $\mu$ be a Radon measure on $\mathbb{R}^d$, let $A\subset\mathbb{R}^d$ be a set, and let $\mathcal{B}$ be a family of balls such that each point of $A$ is the center of balls in $\mathcal{B}$ of arbitrarily small radius. Then there is pairwise disjoint collection $\{ B_i\}\subset\mathcal{B}$ such that $\mu(A\backslash \bigcup_i B_i)=0$. For a proof, see e.g. Mattila's book, Theorem 2.8.

With this theorem it is easy to prove the following: if $\mu$ is a Radon measure on $\mathbb{R}^d$, then
$$
\liminf_{r\to 0}\frac{\mu(B(x,r))}{r^d} > 0 \quad \text{for } \mu\text{-a.e. } x.
$$

To prove this, suppose $\mu$ is a measure such that the claim fails. We may assume without loss of generality that $\mu$ has bounded support (otherwise, let $\mu_N$ be the restriction of $\mu$ to $B(0,N)$; the theorem holds for all $\mu_N$ so it also holds for $\mu$).

Fix a small $\varepsilon>0$. Let $\mathcal{B}$ be the family of all balls $B$ of radius $r\le 1$ such that $\mu(B) < \varepsilon r^d$. Moreover, let $A$ be the set of all $x$ such that $\mu(B(x,r_i)) < \varepsilon r_i^d$ for a sequence $r_i\to 0$. Then $A$ satisfies the hypotheses of Vitali's covering lemma for the family $\mathcal{B}$. Also note that $A\supset A'$, where
$$
A' = \{x\in\mathbb{R}^d: \liminf_{r\to 0}\frac{\mu(B(x,r))}{r^d} = 0\}.
$$
By assumption, $\mu(A')>0$.

Now let $\{ B_i\}$ be the disjoint collection of balls provided by Vitali's covering lemma. On one hand, we have
$$
\mu(\bigcup_i B_i) \ge \mu(A) \ge \mu(A') >0.
$$
On the other hand, by definition of the family $\mathcal{B}$ and the fact that $\{B_i\}$ is disjoint,
$$
\mu(\bigcup_i B_i) = \sum_i \mu(B_i) \le \varepsilon \sum_i r_i^d < C \varepsilon,
$$
where $r_i$ is the radius of $B_i$ and $C$ is a constant that depends only on the size of the support of $\mu$. By taking $\varepsilon$ small enough we obtain a contradiction.

As a corollary, we find that, indeed, it is impossible to have $\mu(B(x,r))\sim r^\alpha$ for any $\alpha>d$.