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.

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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.