Is there a non-quotient stack with affine stabilizers whose good moduli space is a geometric point? Definitions: One says that a map $\pi\colon\mathcal X\to X$ from an algebraic stack to an algebraic space is a good moduli space if $\pi$ is cohomologically affine and universal for maps to schemes. (A good (but not equivalent) way to think about a stack whose good moduli space is a single point is that for any two points, the closure of their orbits intersect.)
One says that a stack $\mathcal X$ is a quotient stack if it is the quotient of an algebraic space by a subgroup of GL_n (see for instance EHKV, which also gives a criterion for a stack to be a quotient stack in terms of vector bundles). 
One says that a stack has the resolution property if every coherent sheaf is a quotient of some vector bundle; 
Totaro's paper The resolution property for schemes and stacks relates the property of being a quotient stack to the resolution property.

Question: Let $\mathcal X$ be a stack with a good moduli space $\mathcal X \to  X$ such that X is a geometric point (i.e., X = Spec k, where k is a separably closed field). Suppose further that the stabilizers are affine linearly reductive groups. Is $\mathcal X$ a quotient stack? 

(See this answer for the definition of stabilizer of a point of a stack that isn't a quotient stack.)
Remarks: 


*

*The condition on stabilizers excludes things like BE with E an elliptic curve.

*The condition that k is separably closed excludes non-trivial gerbes.

*I'd be just as happy with an answer to "Does the resolution property hold for $\mathcal X$ ?". 
 A: This is answered in general by Theorem 13.1 in our paper The étale local structure of algebraic stacks (arXiv:1912.06162). Let $\mathcal{X}$ be a stack with a good moduli space $X$ such that

*

*$\mathcal{X}$ has affine stabilizers,

*$\mathcal{X}$ has separated diagonal, and

*$\mathcal{X}$ is of finite presentation over an algebraic space.

Then $\mathcal{X}$ has the resolution property étale-locally (and even Nisnevich-locally) on $X$. In particular, if $X$ is the spectrum of a field or a henselian local ring, then $\mathcal{X}$ has the resolution property.
Conditions 1 and 2 are necessary. Note that having the resolution property locally on $X$ implies that $\mathcal{X}$ has affine diagonal.
In general $\mathcal{X}$ does not have the resolution property Zariski-locally on $X$. There is an example in SGA 3, Exp X, §1.6 (cf. Remark 2.5 in our paper), of a 2-dimensional torus $G$ over the nodal cubic curve $C$ such that $G$ is not locally isotrivial. This means that $G$ cannot be embedded in $\mathrm{GL}_N$ for any $N$, not even Zariski-locally on $C$. It follows that $\mathcal{X}=BG$ does not have the resolution property Zariski-locally on its good moduli space $X=C$.
