Square differential forms in cohomology

Let $$X$$ be a differentiable manifold (connected, compact, orientiable) of dimension $$4n$$. Consider on $$X$$ a closed $$2n$$-form $$\omega$$, with associated cohomology class $$[\omega] \in H^{2n}(X,\mathbb{R})$$. The integral of its square is some real number, $$\int_X \omega \wedge \omega \in \mathbb{R} \,,$$ which may be negative, positive, or zero. In general, the integrand $$\omega \wedge \omega$$ need not have the same sign everywhere as the result of the integral. However, the integral $$\int_X \omega \wedge \omega$$ is a function only of the cohomology class $$[\omega]$$, while $$\omega \wedge \omega$$ depends on the choice of representative $$\omega \in [\omega]$$. So my question is:

Is it possible to find a cohomologically equivalent $$\omega' \in [\omega]$$ such that everywhere $$\mathrm{sgn}\big(\omega' \wedge \omega'\big) = \mathrm{sgn}\big(\int_X \omega \wedge \omega\big)$$?

In the case that the answer is negative, I wonder if one can give criteria under which it holds.

A negative answer was given to a related question here. However, that answer crucially relied on the existence of the Massey triple product, which vanishes in the present case, so it doesn't seem possible to make a similar argument here.

No, this is not always possible. I'll analyze the case $$n = 1$$ because it is related to symplectic geometry. Note that the sign of $$\omega' \wedge \omega'$$ is not well-defined without choosing a specific orientation on $$M$$ while the condition $$\omega' \wedge \omega' \neq 0$$ does make sense so I'll use this condition instead. A two-form $$\omega' \in \Omega^2(M;\mathbb{R})$$ satisfies $$\omega' \wedge \omega' \neq 0$$ everywhere if and only if $$\omega'$$ is everywhere non-degenerate. So when $$n = 1$$, you are basically asking whether, given a cohomology class $$[\omega]$$ with $$\int_{M} \omega \wedge \omega \neq 0$$, can one find a closed, non-degenerate (i.e symplectic) two-form $$\omega' \in [\omega]$$?
To see that this is not always possible, take for example $$M = \mathbb{CP}^2 \# \mathbb{CP}^2$$. Denote by $$\omega_{FS}$$ the Fubini-Study symplectic form on $$\mathbb{CP}^2$$. Then $$H^2(M) \cong H^2(\mathbb{CP}) \oplus H^2(\mathbb{CP})$$. Let $$\omega \in \Omega^2(M)$$ be some representative of the cohomology class of $$([\omega_{FS}],0)$$. Then by the calculation of the cohomology ring of a connected sum, we have $$\int_{M} \omega \wedge \omega = [\omega] \cdot [\omega] = [\omega_{FS}] \cdot [\omega_{FS}] = \int_{\mathbb{CP}^2} \omega_{FS} \wedge \omega_{FS} > 0.$$
However, it is "well-known" that the manifold $$M$$ has no almost-complex structure and so it has no symplectic two-form in $$[\omega]$$ (or in any other cohomology class).