Let $X$ be a compact Kähler manifold, with fixed Kähler form $\Omega$. Then, the wedge product of two harmonic forms is not necessarily harmonic, as explained for instance here. This prompts the question of whether the product of a $\mathcal{C}^\infty$-antiharmonic form $\alpha \in \mathcal{A}^{p,q}(X)$ with a harmonic form $\beta \in \mathcal{H}^{r,s}(X)$ is antiharmonic. Here, a $\mathcal{C}^\infty$-differential form $\alpha \in \mathcal{A}^{p,q}(X)$ is said to be antiharmonic if $\alpha$ is orthogonal to any harmonic form, with respect to the Hodge inner product $(\alpha,\beta) := \int_X \alpha \wedge (\ast \beta)$, where $\ast$ denotes the Hodge $\ast$-operator. In particular, thanks to the Hodge decomposition, we know that the group of antiharmonic forms coincides with $\mathrm{Im}(d) \oplus \mathrm{Im}(d^\ast)$, where $d$ is the usual exterior differential, and $d^\ast$ is its adjoint with respect to the $\ast$-operator.
Such a statement is clearly true if $\alpha = d(\gamma)$, since the product of a $d$-exact form with a $d$-closed one is again exact, thanks to Leibniz's rule. Therefore, by the Hodge decomposition the only case left to consider is $\alpha = d^\ast(\gamma)$. Is it reasonable to expect that $d^\ast(\gamma) \beta \subseteq \mathrm{Im}(d) \oplus \mathrm{Im}(d^\ast)$, or is this not true?
I would be particularly interested in this property when $p = q = 0$, i.e. when $\alpha = f$ is a function such that $\int_X f \cdot \Omega^n = 0$, where $\Omega \in \mathcal{A}^{1,1}(X)$ is the Kähler form associated to the Kähler metric on $X$. Analogously, one can say that $f = \Delta(g)$ where $g = G(f)$ and $G$ is Green's operator. Therefore, the previous question amounts to ask whether $\int_X \Delta(g) \beta_1 \wedge (\ast \beta_2) = 0$ for every smooth function $g$ and every pair of harmonic forms $\beta_1,\beta_2 \in \mathcal{H}^{p,q}(X)$, where $\ast$ denotes the Hodge $\ast$-operator.
Finally, note that this question is intimately related to the behaviour of the orthogonal projection $H \colon \mathcal{A}^{p,q}(X) \to \mathcal{H}^{p,q}(X)$. More precisely, is it true that $H$ is a map of $\mathcal{C}^\infty(X)$-modules, i.e. that $H(f \alpha) = H(f) H(\alpha)$ for every smooth function $f$ and every $\alpha \in \mathcal{A}^{p,q}(X)$? Note that $H(f)$ is a constant, so this property might be too much to be true. Moreover, note that an analogous statement is not true in general if $f$ is replaced by a differential form of higher degree, unless we are in a situation where the wedge product of two harmonic differential forms is harmonic.
I don't think this is true in general. At least it's not true for your restated claim, that $\int_X f \beta_1 \wedge *\beta_2 = 0$ for harmonic forms $\beta_j$ and smooth functions $f$ such that $\int_X f \omega^n = 0$.
Consider a curve $X$ of genus greater than 1 and a metric $\omega$ that has constant curvature and volume 1. Let $\sigma$ be a nonzero holomorphic $1$-form on $X$ and note that $\sigma$ is harmonic. We can write $\sigma \wedge * \sigma = \sigma \wedge \overline \sigma = g \omega$, where $g \geq 0$ is a smooth real-valued function. (Maybe I lost a sign here, but it's not important.)
Let $c = \int_X g \omega = |\sigma|^2$ and let $f = g - c$, so $f$ has an average of zero over $X$. Then $$ \int_X f \sigma \wedge *\sigma = \int_X fg \omega = \int_X g^2 \omega - \biggl(\int_X g\omega\biggr)^2 = |g|^2 - \langle g, 1 \rangle^2. $$ This is zero if and only if $g$ is a multiple of the constant function 1 by Cauchy-Schwarz. But the holomorphic 1-form $\sigma$ is not zero, but has a zero somewhere because $X$ has genus greater than one, so $g$ is not constant.