Can a class be represented by both a $(p,q)$-form and a $(p',q')$-form? Suppose $X$ is a complex manifold. 
If $X$ is Kähler, the cohomology groups decompose into subgroups represented by $(p,q)$-forms. 
If $X$ is not Kähler, I think the decomposition may not hold? 
Is there an example where we have a nonzero class be represented by both a $(p,q)$-form and a $(p',q')$-form with $(p, q) \neq (p',q')$?
 A: This is meant as a supplement to Michael's answer above. Here is an example of a compact complex threefold $X$ and two forms $\gamma_1 \in \mathcal{E}^{p,q}$ and $\gamma_2 \in \mathcal{E}^{p',q'}$ such that $(p,q) \neq (p',q')$ and $[\gamma_1] = [\gamma_2] \in H_{dR}^2(X)$, but also $[\gamma_1] \neq 0$. This will make use of results found in Cordero, Fernández, Gray, The Frölicher spectral sequence for compact nilmanifolds.
Applying Theorem 6 in this paper gives us that there exists a simply connected 6-dimensional real Lie group with a left-invariant complex structure whose left-invariant $(1,0)$-forms are spanned by $\omega_1, \omega_2, \omega_3$ and whose left-invariant $(0,1)$-forms are spanned by $\bar \omega_1, \bar \omega_2, \bar \omega_3$, with structure equations given by \begin{align*} d\omega_1 &= 0, \ d\bar \omega_1 = 0, \\ d\omega_2 &= 0, \ d\bar \omega_2 = 0, \\ d\omega_3 &= \omega_1\omega_2 - \bar \omega_1 \omega_2, \ d\bar \omega_3 = \bar \omega_1 \bar \omega_2 - \omega_1 \bar \omega_2. \end{align*} Since the (dual) structure constants are rational, a theorem of Malcev (see Corollary 8 in the paper) tells us that there exists a discrete subgroup $\Gamma$ in $G$ such that $\Gamma \backslash G$ is a closed 6-manifold. The forms $\omega_i$ and $\bar \omega_i$ descend to $\Gamma \backslash G$ along with the complex structure on $G$. A theorem of Nomizu (referenced in this context in Hasegawa, Minimal models of nilmanifolds  at the end of Section 1) tells us that the de Rham cohomology of $\Gamma \backslash G$ is isomorphic to to the cohomology of the differential graded algebra generated in degree one by $\{\omega_1, \omega_2, \omega_3, \bar \omega_1, \bar \omega_2, \bar \omega_3\}$ with the differential $d$ prescribed above. 
Now, take the $(2,0)$-form $\omega_1\omega_2$ and the $(1,1)$-form $\bar \omega_1 \omega_2$. Note that they are both closed, and their difference is $d$-exact (since $d\omega_3 = \omega_1\omega_2 - \bar \omega_1 \omega_2$). So, $[\omega_1\omega_2] = [\bar \omega_1 \omega_2]$ in de Rham cohomology. We see that $\omega_1\omega_2$ is not in the image of $d$, that is $[\omega_1\omega_2] \neq 0$. 
A: I think you want an example of a compact complex manifold $X$ and differential forms $\gamma \in \mathcal{E}^{p,q}(X)$ and $\gamma'\in \mathcal{E}^{p',q'}(X)$ with $(p',q') \neq (p, q)$ such that $[\gamma] = [\gamma']$ in de Rham cohomology.
Let $X$ be a compact complex three-dimensional manifold with a non-closed holomorphic one-form $\alpha$, i.e. $\alpha \in \mathcal{E}^{1,0}(X)$ such that $\overline{\partial}\alpha = 0$, but $d\alpha = \partial\alpha \neq 0$. An example of such a manifold is the Iwasawa manifold, see below for details.
Let $\gamma = \partial\alpha \in \mathcal{E}^{2,0}(X)$ and $\gamma' = \overline{\gamma} \in \mathcal{E}^{0,2}(X)$. Note that 
$$d\gamma = \partial\gamma + \overline{\partial}\gamma = \partial\partial\alpha + \overline{\partial}\partial\alpha = -\partial\overline{\partial}\alpha = 0$$
and $d\gamma' = d\overline{\gamma} = \overline{d\gamma} = 0$, so $\gamma$ and $\gamma'$ define de Rham cohomology classes $[\gamma], [\gamma'] \in H^2_{\text{dR}}(X, \mathbb{C})$ which satisfy
$$[\gamma] = [\partial\alpha] = [d\alpha] = 0 = [d\bar{\alpha}] = [\overline{d\alpha}] = [\overline{\partial\alpha}] = [\overline{\gamma}] = [\gamma'].$$
This is somewhat unsatisfying. It would be interesting to see an example where $[\gamma] = [\gamma'] \neq 0$, but I have not been able to construct one yet.

On a compact Kähler manifold, every holomorphic form is closed. Without the Kähler hypothesis, one can still show that on a compact complex $n$-dimensional manifold, every holomorphic $(n-1)$-form is closed. Therefore, the first possible example of a non-closed holomorphic form is a holomorphic one-form on a three-dimensional compact complex manifold. Such examples exist.
Let $R$ be a commutative ring with identity. The three-dimensional Heisenberg group over $R$ is
$$\mathbb{H}(3, R) = \left\{\begin{bmatrix} 1 & z^1 & z^3\\ 0 & 1 & z^2\\ 0 & 0 & 1\end{bmatrix} : z^1, z^2, z^3 \in R\right\}.$$
The Iwasawa manifold $\mathbb{I}_3$ is the quotient of $\mathbb{H}(3, \mathbb{C})$ by the discrete subgroup $\mathbb{H}(3, \mathbb{Z}[i])$ acting on the left, i.e. $\mathbb{I}_3 := \mathbb{H}(3, \mathbb{Z}[i])\setminus\mathbb{H}(3, \mathbb{C})$. More precisely, for $A \in \mathbb{H}(3, \mathbb{Z}[i])$, $\gamma_A : \mathbb{H}(3, \mathbb{C}) \to \mathbb{H}(3, \mathbb{C})$ given by $\gamma_A(M) = AM$ is a biholomorphism and $\Gamma = \{\gamma_A \mid A \in \mathbb{H}(3, \mathbb{Z}[i])\}$ acts properly discontinuously on $\mathbb{H}(3, \mathbb{C})$, so the quotient $\mathbb{I}_3$ is a complex manifold. Moreover, the quotient is compact.
Let 
$$A = \begin{bmatrix} 1 & w^1 & w^3\\ 0 & 1 & w^2\\ 0 & 0 & 1\end{bmatrix} \in \mathbb{H}(3, \mathbb{Z}[i]).$$
The differential forms $dz^1$, $dz^2$, and $dz^3-z^1dz^2$ on $\mathbb{H}(3, \mathbb{C})$ form a basis of holomorphic one-forms. As
$$\gamma_A^*(dz^1) = d(z^1 + w^1) = dz^1,$$ 
$$\gamma_A^*(dz^2) = d(z^2 + w^2) = dz^2,$$
and
\begin{align*}
\gamma_A^*(dz^3-z^1dz^2) &=  d(z^3 + w^1z^2 + w^3) - (z^1 + w^1)d(z^2 + w^2)\\ 
&= dz^3 + w^1dz^2 - z^1dz^2 - w^1dz^2\\ 
&= dz^3 - z^1dz^2,
\end{align*}
the forms $dz^1, dz^2, dz^3-z^1dz^2$ are left-invariant, so they descend to holomorphic one-forms $\alpha^1, \alpha^2, \alpha^3$ on $\mathbb{I}_3$. As $dz^1$ and $dz^2$ are closed, $\alpha^1$ and $\alpha^2$ are closed. However, 
$$d(dz^3 - z^1dz^2) = -dz^1\wedge dz^2$$
so $d\alpha^3 = -\alpha^1\wedge\alpha^2 \neq 0$. Therefore $\alpha^3$ is a holomorphic one-form which is not closed. 
One can deduce a few more things about $\mathbb{I}_3$ from the above. First of all, the forms $\alpha^1, \alpha^2, \alpha^3$ form a basis for the holomorphic one-forms on $\mathbb{I}_3$, so $h^{1,0}(\mathbb{I}_3) = 3$. Secondly, as $\mathbb{I}_3$ admits a holomorphic form which is not closed, it cannot admit a Kähler metric.
