I'm not a geometric measure theorist, so this answer is just my approach for this problem, which is rather hands-on. The question is equivalent to verifying $I(f) = J(f)$ for all $f \in C_c(\mathbb{R}^d)$, where $I(f) = \int_{\mathbb{R}^d} f \, d\mu$, $J(f) = \int_{\mathbb{R}^d} f \, d\nu$; recall that the action of the integration functional on the continuous, compactly-supported functions suffices to recover a Radon measure completely.
First, take any nonnegative continuous function $\phi : \mathbb{R}^d \to \mathbb{R}$ that is radially-decreasing, and supported in $B(0, R)$ for some $R > 0$. By using the profile of $\phi$, we may approximate with a weighted sum of indicator functions of balls centered at the origin, $\sum_{i = 1}^{n} a_i \chi_{B(0, r_i)}$ for $a_i \geq 0$, $0 \leq r_i \leq R$. Moreover, this approximation can clearly be made uniformly close (partitioning the ball into sufficiently-fine annuli).
Since $\mu(B(0, r_i)) = \nu(B(0, r_i))$ by hypothesis, the integrals of the radial step functions $\sum_{i = 1}^{n} a_i \chi_{B(0, r_i)}$ are identical with respect to either $\mu$ or $\nu$. Then taking limits, we find that $\int \phi \, d\mu = \int \phi \, d\nu$ for any $\phi$ of this form. Finally, we see this logic works for balls centered at points other than the origin. So actually, any compactly-supported nonnegative continuous $\phi$, that is radially-decreasing about some point $y \in \mathbb{R}^d$, will obey the relation $\int_{\mathbb{R}^d} \phi \, d\mu = \int_{\mathbb{R}^d} \phi \, d\nu$.
We now take an arbitrary function $f \in C_c(\mathbb{R}^d)$. Letting $\phi_{\epsilon}$ be a standard mollifier, we have $\phi_{\epsilon} * f \to f$ uniformly on $\mathbb{R}^d$ as $\epsilon \to 0$, and if $\phi \in C_c$ as well, we can ensure the $\phi_{\epsilon} * f$ for $\epsilon > 0$ small also share a common support. It is easy to show, for any fixed $\epsilon > 0$, the convolution itself can be approximated as $$(\phi_{\epsilon} * f)(x) \approx \sum_i \phi_{\epsilon}(x - y_i) f(y_i) |C_i|$$ for a finite collection of disjoint dyadic cubes $C_i$ covering the support of $f$, interior sample points $y_i \in C_i$, and where $| \cdot |$ denotes the Lebesgue measure. This approximation can be made uniform over all $x \in \mathbb{R}^d$. (This is essentially just an exercise in applying the uniform continuity and boundedness of $f$ and $\phi_{\epsilon}$.)
Summing up, we can first approximate $\int f \, d\mu$ and $\int f \, d\nu$ to error, say, $< \eta / 2$ by replacing them with $\int \phi_{\epsilon_0} * f \, d\mu$ and $\int \phi_{\epsilon_0} * f \, d\nu$, when $0 < \epsilon_0 \ll 1$ is sufficiently small. We can then use the Riemann-style sums to approximate the convolution $\phi_{\epsilon_0} * f$ pointwise with $\sum_i \phi_{\epsilon_0}(\cdot - y_i) f(y_i) |C_i|$, which contributes another error of $< \eta / 2$ in the integrals if we take our dyadic cubes $C_i$ to all have sufficiently-small side length (depending on the parameter $\eta$, the functions $f$, $\phi_{\epsilon_0}$, the $\mu$- and $\nu$-measures of their supports, etc.).
But from our calculation in the radially-decreasing case, $\int \phi_{\epsilon}(x - y) \, d\mu_x = \int \phi_{\epsilon}(x - y) \, d\nu_x$ for any choice of $y \in \mathbb{R}^d$, $\epsilon > 0$. By linearity, the $\mu$ and $\nu$ integrals of the function $\sum_i \phi_{\epsilon_0}(\cdot - y_i) f(y_i) |C_i|$ are equal.
In total, we see the integrals $\int f \, d\mu$ and $\int f \, d\nu$ are within $\eta / 2 + \eta / 2 = \eta$ of each other, where $\eta > 0$ was arbitrary. Hence $I(f) = J(f)$, and so the Radon measures $\mu$ and $\nu$ must coincide.
