Unconditional? Certainly not. E.g., suppose that $\mu$ is the standard Gaussian measure on $\mathbb R^d$ and $B$ is the ball of radius $r>0$ centered at $0$. Then for any $\delta\in(0,1)$, by the law of large numbers,
$$1-\mu(B^\epsilon)=P\Big(\frac1d\sum_1^d Z_i^2>\frac{(r+\epsilon)^2}d\Big)\to1$$
for $d\to\infty$ and $\epsilon\in(0,(1-\delta)\sqrt d-r\,]$, where $Z_1,Z_2,\dots$ are iid $N(0,1)$. So, asymptotically there is no decrease at all in $\epsilon\le(1-\delta)\sqrt d-r$. So, any bound of the form $1-\mu(B^\epsilon)\le\epsilon\,\text{poly}(\epsilon)$ will not hold in this example.