When $(X,g,\omega)$ is a compact Kaehler surface, the integral classes representable as the real part of a (2,0)-form can be represented as the curvatures of anti-instantons.
One has two splittings of the $g$-harmonic 2-forms $\mathcal{H}^2_g(X)$: the Hodge-theoretic one, and the one determined by the Hodge star, $$ \mathcal{H}^2_g(X) = \mathcal{H}^+_g(X) + \mathcal{H}^-_g(X), $$ into self-dual and anti-self-dual harmonic forms.
The relation with the Hodge decomposition is that $$ \mathcal{H}^+ = \mathbb{R}\omega \oplus (\mathcal{H}^{2,0}\oplus \mathcal{H}^{0,2})_{\mathbb{R}} $$ and $$ \mathcal{H}^- = (\mathcal{H}^{1,1}_0)_{\mathbb{R}}$$ (trace-free part of the real (1,1)-forms). We can check this pointwise, where it's linear algebra. The wonderful thing about Hodge theory is that it then implies the corresponding cohomology-level statement. (Note that $\mathcal{H}^\pm$ are maximal positive-definite (+) and negative-definite (-) subspaces of the wedge-product form on $H^2(X;\mathbb{R})$. So as a by-product we get the Hodge index theorem.)
Suppose that there's an integral class $c$ in $ (\mathcal{H}^{2,0}\oplus \mathcal{H}^{0,2})_{\mathbb{R}}$. It's represented by a hermitian line bundle $L_c$, and (by Hodge theory again) we can choose a unitary connection in $L_c$ whose curvature is harmonic and hence self-dual.
Corrected: This connection, which is called an abelian anti-instanton, is not quite unique, because we could add to it $i$ times any closed 1-form; but if this closed form is exact, or represents $2\pi$ times an integral cohomology class, then the new connection is gauge-equivalent to the original one. Hence, the space of anti-instantons mod gauge is an torsor for the torus $H^1(X;S^1)$. The catch, I think, is that for generic Kaehler metrics in a fixed Kaehler class, there will be no such integral classes unless $\mathcal{H}^{1,1}_0=0$.