I was wondering whether there is a version of Hironaka's resolution of singularities theorem that applies to stacks. Specifically, I was wondering about the following situation:
We define the quotient stack $S = \mathbb{A}^n/\mathbb{G}_{\operatorname{m}}$, where the multiplicative group acts on $\mathbb{A}^n$ by scaling. In other words, $S$ is just projective $(n-1)$-space, but with the image $\mathbf{O}$ of the origin left in. Now $S$ presumably has a bad singularity at $\mathbf{O}$. Does there exist a smooth stack $\widetilde{S}$ and a "resolution of singularities" morphism $f:\widetilde{S} \to S$?
Note that I am being deliberately vague about the nature of $f$. In the context of schemes, I would for example have required $f$ to be birational, however I am not sure that this makes sense for stacks.
First I want to point out that the stack $S$ in your question $\textbf{is}$ a smooth stack over the base field. It has a smooth cover $\mathbb{A}^n \to S$ given by the quotient map and $\mathbb{A}^n$ is smooth so $S$ is smooth since the property of being smooth is local on the source in the smooth topology. The problem with the stack $S$ at the origin is that it is not $\textbf{separated}$.
As for your original question, I believe the answer is yes: finite type reduced algebraic stacks over a field of characteristic zero should admit resolution of singularities.
For Deligne-Mumford stacks this follows from equivariant resolution of singularities for a variety with a finite group action (see for example https://arxiv.org/abs/alg-geom/9609013). In the general case you can use one of the algorithms that produce a "canonical" resolution such as the one described in the following paper:
Encinas, S.; Villamayor, O. (1998), "Good points and constructive resolution of singularities.", Acta Math. 181 (1): 109–158, doi:10.1007/BF02392749, MR1654779
In particular, their resolution is functorial with respect to smooth morphisms so given a smooth atlas $R \rightrightarrows U$ of an algebraic stack $S$, the functorial resolutions $f: U' \to U$ and $g: R' \to R$ give you a smooth groupoid $R' \rightrightarrows U'$ defining an algebraic stack $S' \to S$. Since $U'$ is smooth over the base field, so is $S'$. Furthermore, $f$ and $g$ are isomorphisms over the (dense open) locus where $U$ us smooth, so the map $S' \to S$ is a resolution of singularities.