I think that the answer is yes. Assume by contradiction that the support $S$ of $\phi$ is nonempty. Since $S$ is closed, it is in particular a complete metric space.

Let $X:=\{\psi\in C^\infty(\mathbb{R}^d):\psi\equiv 0\text{ on }\mathbb{R}^d\setminus B_1\}$, which is a Fréchet space.
Apply now, for any fixed $x\in S$, the uniform boundedness principle to the family of functionals
$$X\to\mathbb{R},\qquad\{\psi\mapsto\langle\phi, \psi^{\lambda}_x \rangle\mid 0<\lambda<1\},$$ obtaining that for some minimal $k(x)\in\{1,2,\dots\}$ it holds
$$ |\langle\phi,\psi_x^\lambda\rangle|\le k(x)\|\psi\|_{C^{k(x)}} $$
(for any $0<\lambda<1$). The sets $S_N:=\{x\in S:k(x)\le N\}$ are closed and cover $S$, so one of them has nonempty interior: say that $B_r(x)\cap S\subseteq S_N$ for some $x\in S$.

Now, for any positive integer $M$ large enough, we can easily find a collection $(\rho_{M,j})_j\subseteq C^\infty_c(B_r(x))$, with cardinality $O(2^{Md})$, of functions whose supports have diameter less than $2^{-M}$ and satisfying $\|\rho_{M,j}(2^{-M}\cdot)\|_{C^N}=O(1)$, as well as $\sum_j\rho_{M,j}\equiv 1$ on $B_{r/2}(x)$ (say).

Thus, for any $\psi\in C^{\infty}_c(B_{r/2}(x))$,
$$ \langle\phi,\psi\rangle=\sum_j\langle\phi,\rho_{M,j}\psi\rangle $$
and we can discard all the terms where $\rho_{M,j}$ has support disjoint from $S$. As for the other terms,
let $y_j\in S\cap\text{supp }\rho_{M,j}$; notice that
$$\langle\phi,\rho_{M,j}\psi\rangle=2^{-Md}\langle\phi,\eta_{y_j}^{2^{-M}}\rangle$$
where $\eta:=\rho_{M,j}(y_j+2^{-M}\cdot)\psi(y_j+2^{-M}\cdot)$. Thus,
$$|\langle\phi,\rho_{M,j}\psi\rangle|\le 2^{-Md}N\|\rho_{M,j}(y_j+2^{-M}\cdot)\psi(y_j+2^{-M}\cdot)\|_{C^N}$$
$$\le CN 2^{-Md}\|\psi(y_j+2^{-M}\cdot)\|_{C^N}. $$
Summing in $j$ and letting $M\to\infty$, this easily leads to the bound
$$ |\langle\phi,\psi\rangle|=O(\|\psi\|_\infty), $$
showing that $\phi$ restricts to a measure on $B_{r/2}(x)$. The hypothesis, plus well-known differentiation theorems, imply that $\phi\equiv 0$ here (see below), contradicting the fact that $x\in S$.

**Addendum.** Proof of the fact that $\phi$ vanishes on $B_r(x)$:
Let us restrict our attention to $B_r(x)$. We split the measure $\phi=\phi^+-\phi^-$ into positive and negative part and we set $\mu:=\phi^-+\mathcal{L}^d$, $\nu:=\phi^+$. Assume that some $y\in B_r(x)$ satisfies
$$\lim_{\rho\to 0}\frac{\nu(B_\rho(y))}{\mu(B_\rho(y))}=+\infty.$$
Pick any radial $\psi\in C^\infty_c$
with $\psi\ge 0$ and $\int_{\mathbb{R}^d}\psi\,d\mathcal{L}^n=1$. Notice that
$$\frac{\int\psi_y^\lambda\,d\nu}{\int\psi_y^\lambda\,d\mu}=\frac{\int_0^\infty\nu(\{\psi_y^\lambda>t\})\,dt}{\int_0^\infty\mu(\{\psi_y^\lambda>t\})\,dt}\to+\infty$$ as $\lambda\to 0$ (since the superlevels are smaller and smaller balls). Thus,
$$\frac{\int\psi_y^\lambda\,d\phi}{\int\psi_y^\lambda\,d\mu}\ge\frac{\int\psi_y^\lambda\,d\nu}{\int\psi_y^\lambda\,d\mu}-1\to+\infty,$$
but $\int\psi_y^\lambda\,d\mu\ge\int\psi_y^\lambda\,d\mathcal{L}^d=1$, contradicting the hypothesis. So such $y$ does not exist and by Theorem 2.22 (Besicovitch derivation theorem) in *Ambrosio, Fusco, Pallara, Functions of Bounded Variation and Free Discontinuity Problems* we have $\nu\ll\mu$.

Since $\phi^+$ and $\phi^-$ are mutually singular, we deduce $\phi^+\ll\mathcal{L}^d$.
Repeating all the arguments with $-\phi$ instead of $\phi$, we get $\phi^-\ll\mathcal{L}^d$ as well.
So $\phi$ has a density with respect to $\mathcal{L}^d$. For all $y$ such that $\lim_{\rho\to 0}\frac{\phi(B_\rho(y))}{\mathcal{L}^d(B_\rho(y))}$ exists finite, the ratio $\frac{\int\psi_y^\lambda\,d\phi}{\int\psi_y^\lambda\,d\mathcal{L}^d}$ converges to the same limit. Applying the aforementioned theorem and the hypothesis, we obtain that the density is zero.