This doesn't answer the original question; instead, it argues that the prohibition against using any derivatives of the 'candidate' curvature form is not really necessary. The first step is to show that, if a continuous $\frak{g}$-valued $2$-form $\Phi$ is the curvature of a continuous $\frak{g}$-valued $1$-form $\alpha$ (which must necessarily have a continuous exterior derivative), then $\Phi$ itself must have a continuous exterior derivative and the Bianchi identity must hold. (Note that this is weaker than claiming that $\Phi$ is differentiable, which might well not be true. It's easy for a differential form to be only $C^0$ but still have a continuous exterior derivative.)
I'm going to concentrate on the 'local' problem, i.e., I'm going to assume that the bundle is trivial. Thus, suppose that you have a continuous $2$-form $\Phi$ on a manifold $M$ that takes values in a Lie algebra $\frak{g}$. We want to know whether there is a $\frak{g}$-valued $1$-form $\alpha$ that is continuous (and has a continuous exterior derivative $d\alpha$) such that $\Phi = d\alpha + \frac12[\alpha,\alpha]$.
Let $\mathcal{C}_p(M)$ denote the vector space of smooth singular $p$-chains in $M$ with real coefficients, and let $\partial:\mathcal{C}_p(M)\to \mathcal{C}_{p-1}(M)$ denote the boundary operator. If the desired $\alpha$ exists, then, for every $\sigma\in\mathcal{C}_3(M)$, one has the identity
$$
\int_{\partial\sigma}\Phi = \int_{\partial\sigma} d\alpha + \frac12[\alpha,\alpha] = \int_{\partial\sigma}\frac12[\alpha,\alpha]= \int_{\sigma} [d\alpha,\alpha]=\int_{\sigma} [\Phi,\alpha].
$$
Note that I did not differentiate $\Phi$. This calculation shows that, if $\alpha$ exists, then $\Phi$ has to have a continuous exterior derivative, in the sense that there has to be a continuous $3$-form $\Psi$ such that
$$
\int_{\partial\sigma}\Phi = \int_\sigma \Psi
$$
for all $\sigma\in\mathcal{C}_3(M)$. Thus, a necessary condition that $\Phi$ be the curvature of a continuous connection form $\alpha$ having a continuous exterior derivative is that $d\Phi$ must exist and be continuous.
Moreover, this calculation shows that, if there is a continuous solution $\alpha$, it must satisfy the linear algebraic equation $[\Phi,\alpha] = d\Phi$. In particular, $d\Phi$ must lie in the image of the operator $W_\Phi: {\frak{g}}\otimes\Omega^1(M)\to
{\frak{g}}\otimes\Omega^3(M)$ defined by $W_\Phi(\alpha) = [\Phi,\alpha]$. This restriction frequently gives us algebraic equations that $(\Phi,d\Phi)$ must satisfy.
When ${\frak{g}} = {\frak{su}}(2)$ and the dimension of $M$ is $4$, the map $W_\Phi$ is an isomorphism for 'generic' $\Phi$ (where 'genericity' is a pointwise condition). In such a case, there will exist a unique $\alpha$ satisfying $[\Phi,\alpha] = d\Phi$ (and it will necessarily be continuous). If this $\alpha$ does not have a continuous exterior derivative or, if does have a continuous exterior derivative but doesn't satisfy $d\alpha + \frac12[\alpha,\alpha] = \Phi$, then there is no solution. Thus, in this `generic' case, one sees that the condition to be a curvature is that certain combinations of the first and second derivatives of $\Phi$ exist and provide a solution to the problem, just by differentiation.
When ${\frak{g}} = {\frak{su}}(2)$ and the dimension of $M$ is greater than $4$, the map $W_\Phi$ is an injection for 'generic' $\Phi$ (where 'genericity' is a pointwise condition), but it is not a surjection, so the condition $[\Phi,\alpha] = d\Phi$ already places first-order conditions on $\Phi$ and uniquely determines the only possible candidate $\alpha$, when it exists at all.
Other Lie algebras have to be treated on a case-by-case basis, to some extent, though there is probably something similar to the above story for all semi-simple Lie algebras.
