Case 1: When the domain of interest is the whole $\mathbb{R}^3$, the answer is yes. This is obtained by the Helmholtz decomposition for $C^1$-vector field:
$$
\mathbf{w} = \nabla \phi + \nabla \times \mathbf{A} ,\tag{1}
$$
where
$$
\phi(x) = -\int_{\mathbb{R}^3} \Phi(y-x) \nabla_y \cdot \mathbf{w} (y) \,dy,
\\
\mathbf{A} (x) = \int_{\mathbb{R}^3} \Phi(y-x) \nabla_y \times \mathbf{w} (y)\,dy .
$$
The $\Phi(\xi)$ is the fundamental solution to $-\Delta \Phi = \delta_0$ in whole $\mathbb{R}^3$.
Here we have to assume $\mathbf{w}$ has certain decaying properties so that the surface integral $\displaystyle\int_{\partial B} \Phi(y-x)\mathbf{n}\times \mathbf{w} (y)\,dS(y)$ and $\displaystyle\int_{\partial B} \Phi(y-x)\mathbf{n}\times \mathbf{w} (y)\,dS(y)$ vanish for arbitrary large ball $B$, $|\mathbf{w}|\to 0 $ when $|y|\to \infty$ suffices.
Now if $\mathbf{w} $ is divergence free, we can see that $\mathbf{w} = \nabla \times \mathbf{A}$.
Case 2: When the domain of interest $\Omega$ is simply-connected, bounded and open, we have to consider the boundary condition of $\mathbf{w}$. The potentials in (1) will become:
$$
\phi(x) = -\int_{\Omega} \Phi(y-x) \nabla_y \cdot \mathbf{w} (y) \, dy +\int_{\partial \Omega}\Phi(y-x)\mathbf{w}(y)\cdot \mathbf{n}\,dS(y),
\\
\mathbf{A}(x) = \int_{\Omega} \Phi(y-x) \nabla_y \times \mathbf{w}(y)\,dy - \int_{\partial \Omega} \Phi(y-x)\mathbf{n}\times \mathbf{w}(y)\,dS(y).
$$
Here $\Phi$ is the fundamental solution or Green function for
$$
\left\{\begin{aligned}
-\Delta_y \Phi(y-x) &= \delta_x(y) \quad \text{in }\Omega,
\\
\Phi &=0 \quad \text{on } \partial\Omega.
\end{aligned}\right.
$$
We have the boundary term here, by divergence theorem:
$$
\int_{\partial \Omega}\Phi(y-x)\mathbf{w}(y)\cdot \mathbf{n}\,dS(y)
= \int_{\Omega}\nabla \cdot\Big(\Phi(y-x)\mathbf{w}(y)\Big)\,dy
\\
= \int_{\Omega} \Phi(y-x) \nabla_y \cdot \mathbf{w} (y) \, dy
+ \int_{\Omega} \nabla_y \Phi(y-x) \cdot \mathbf{w} (y) \, dy= \int_{\Omega} \nabla_y \Phi(y-x) \cdot \mathbf{w} (y) \, dy.
$$
Hence we can see that in this case, $\mathbf{w} = \nabla \times \mathbf{A}$ if
$\mathbf{w}$ is perpendicular to the gradient field of the fundamental solution pointwisely (or a.e.) in this domain.
$\mathbf{w}\cdot \mathbf{n}=0$ pointwisely (or a.e.) on the boundary of this domain.
Counterexamples for not making boundary term vanish: in $\mathbb{R}^3$, let $\phi$ be the solution to the following boundary value problem
$$
\left\{\begin{aligned}
-\Delta \phi&= 0\quad \text{in }\Omega,
\\
\phi &=g \neq 0 \quad \text{on } \partial\Omega.
\end{aligned}\right.
$$
We can see that if we let $\mathbf{w} = \nabla \phi$, i.e., the gradient of a harmonic function, $\mathbf{w} $ has zero divergence, yet $\mathbf{w} $ is never a curl field.
This is the classical case, for vector potentials (or Helmholtz decomposition) in the distributional sense, this is my favorite reference: http://perso.univ-rennes1.fr/monique.dauge/publis/ABDG_VPot.pdf
