Edited: One precise result is as follows, from Evans' Weak Convergence Methods for Nonlinear Partial Differential Equations (AMS 1990). We modify the proof of Theorem 6 (Compactness for Measures), page 7, to reach the case $U = \mathbb{R}^d$.
Theorem: Let $(\mu_n)_{n \in \mathbb{N}}$ be a tight sequence in $\mathcal{M}(\mathbb{R}^d)$ converging weakly to $\mu$. Then $(\mu_n)_{n \in \mathbb{N}}$ is convergent in $W^{- 1, q}(\mathbb{R}^d)$ for $1 \leq q < \frac{d}{d - 1}$.
The argument is as follows: take and fix any smooth, nonnegative, radially-decreasing cutoff with $\varphi(x) = 1$ for $|x| \leq 1$, $\text{supp}(\varphi) \subseteq B(0, 2)$.
Then for any $R > 0$, the map $f \mapsto f \varphi_R$ for $\varphi_R(x) = \varphi(x / R)$ gives a bounded linear operator $T_R : W^{1, q'}(\mathbb{R}^d) \to W^{1, q}_0(B(0, 2 R))$, and it can be verified from the Leibnitz rule that $\|T_R\|_* \leq 2 + R^{- 1} \|\nabla \varphi\|_{L^{\infty}(\mathbb{R}^d)}$, and thus $\|T_R\|_* \leq 3$ for $R > 1$ sufficiently large.
Letting $\epsilon > 0$, we use tightness to take $R^* \gg 1$ such that for $B = B(0, R^*)$, $|\mu_n|(B^c) < \epsilon$ for all $n$, and also $|\mu|(B^c) < \epsilon$. Increasing $R^*$ if necessary, we may ensure $\|T_{R^*}\|_* \leq 3$.
We let $q' = (1 - \frac{1}{q})^{- 1}$, and form $W^{1, q'}_0(2 B)$. Because $q < \frac{d}{d - 1}$, $q' > d$, and so $W^{1, q'}_0(2 B) \hookrightarrow C_0^{1 - d / q'}(2 B)$ by the standard Sobolev-Morrey embedding result.
By Arzelà-Ascoli and the boundedness of the domain $B$, $C_0^{1 - d / q'}(2 B) \hookrightarrow C_0(2 B)$ compactly. It follows that the image of any norm-bounded set $S$ from $W^{1, q'}_0(2 B)$, $\iota(S) \subseteq C_0(2 B)$, is precompact.
So for the unit ball of $W^{1, q'}(\mathbb{R}^d)$, if we map it by $T_{R^*}$ to a bounded subset of $W^{1, q'}_0(2 B)$, this will be a totally-bounded subset in the space $C_0(2 B)$. Thus there is a finite collection $(g_i)_{i = 1}^{N(\epsilon)} \subseteq C_0(2 B)$ of points that produces an $\epsilon$-net of the image.
Now for any $f \in W^{1, q'}(\mathbb{R}^d)$ with $\|f\|_{W^{1, q'}(\mathbb{R}^d)} = 1$, we estimate $$\left| \int_{\mathbb{R}^d} f \, d\mu_n - \int_{\mathbb{R}^d} f \, d\mu \right| \leq \left| \int_{\mathbb{R}^d} \varphi_{R^*} f \, d\mu_n - \int_{\mathbb{R}^d} \varphi_{R^*} f \, d\mu \right| + \left| \int_{\mathbb{R}^d} (1 - \varphi_{R^*}) f \, d\mu_n - \int_{\mathbb{R}^d} (1 - \varphi_{R^*}) f \, d\mu \right|.$$
For the second term, we consider the support of the integrand, and use a base-height bound: $$(\text{II}) \leq \|(1 - \varphi_{R^*}) f\|_{L^{\infty}(\mathbb{R}^d)} (|\mu|(B(0, R^*)^c) + |\mu_n|(B(0, R^*)^c) ),$$ and since we know $|\mu_n|(B^c) < \epsilon$ and likewise for $\mu$, this is $$\leq C_{d, q'} \|(1 - \varphi_{R^*}) f\|_{W^{1, q'}(\mathbb{R}^d)} (\epsilon + \epsilon) \leq 2 \epsilon C_{d, q'} \|\text{Id} - T_{R^*}\|_{W^{1, q'} \to W^{1, q'}} \leq 2 (1 + 3) \epsilon C_{d, q'},$$ where $C_{d, q'}$ is some universal constant (arising from the Sobolev embedding $W^{1, q'} \hookrightarrow L^{\infty}$), and we recall that $T_{R^*}$ has operator norm $\leq 3$.
It now only remains to deal with the first term. We recall, from the total-boundedness property, that some $g_i \in C_0(2 B)$ will ensure $\|g_i - \varphi_{R^*} f\|_{L^{\infty}(2 B)} < \epsilon$. From this, we can control $$(\text{I}) \leq \left| \int_{\mathbb{R}^d} g_i \, d\mu_n - \int_{\mathbb{R}^d} g_i \, d\mu \right| + \epsilon |\mu_n|(\mathbb{R}^d) + \epsilon |\mu|(\mathbb{R}^d).$$
Weak convergence implies $|\mu_n|(\mathbb{R}^d) \leq M$ uniformly in $n$, and thus this becomes $$\|\mu_n - \mu\|_{W^{- 1, q}(\mathbb{R}^d)} = \sup_{\|f\|_{W^{1, q'}(\mathbb{R}^d)} = 1} |\langle f, \mu - \mu_n \rangle| \leq \sum_{i = 1}^{N} |\langle g, \mu_n - \mu \rangle| + (C_{d, q} + 2 M) \epsilon.$$
As $n \to \infty$, by weak convergence of $\mu_n \rightharpoonup \mu$ and the fact that the $g_i \in C_0(2 B) \subseteq C_0(\mathbb{R}^d)$, we will have $|\langle g_i, \mu_n - \mu \rangle| \to 0$ for every $i \in \{1, \dots, N\}$.
Thus $\limsup_{n \to \infty} \|\mu_n - \mu\|_{W^{- 1, q}(\mathbb{R}^d)} \leq (C_{d, q} + 2 M) \epsilon$, for every $\epsilon > 0$, which means the limit supremum is zero and we have norm convergence.
(It seems like this argument can be modified to give convergence in any $W^{- k, p}(\mathbb{R}^d)$ space with $k$ and $p$ such that $k p' > d$.)
Now, to your question specifically: since you specified the $\mu_n$ as well as the limit $\mu$ are all positive probability measures, we can conclude tightness. Indeed, given any $\delta > 0$, $\mu(\mathbb{R}^d) = 1$ implies that $\mu(K) > 1 - \delta$ for some compact $K$.
Then $\mu_n(K) > 1 - \delta$ for all $n$ sufficiently large, using positivity and weak convergence. Enlarging $K$ if necessary, this shows $\mu_n(K) > 1 - \delta$ for all $n$, and (since $\mu_n(\mathbb{R}^d) = 1$ for each $n$) we get $\mu_n(K^c) = |\mu_n|(K^c) < \delta$, and also $|\mu|(K^c) < \delta$.
This shows tightness of the sequence $(\mu_n)_{n \in \mathbb{N}}$, and thus the result applies.
(We also note that some degree of tightness seems necessary, to obtain norm-convergence in a Sobolev space posed on all of $\mathbb{R}^d$. Otherwise, the sequence $\mu_n = \delta_{n e_1}$ of masses moving to infinity has a weak limit in $(C_0(\mathbb{R}^d))'$ of the $0$ measure. Then $\inf_n \|\mu_n - \mu\|_{W^{- 1, q}(\mathbb{R}^d)} = \inf_n \|\mu_n\|_{W^{- 1, q}(\mathbb{R}^d)} \gtrsim 1$, just by testing against translations of a single function in $W^{1, q'}$.)
