Consider the Hodge numbers
$$h^{p,q}(X) = \dim(H^{p,q}(X)).$$
Then we can ask if the $h^{p,q}$ are topological invariants; if they are not, then clearly there exist homeomorphic complex manifolds $X$ and $Y$ with
$$H^{p,q}(X) \not\cong H^{p,q}(Y).$$
This is an old question, first asked in 1954 by Hirzebruch in the following form:
Are the $h^{p,q}$ and the Chern characteristic numbers of an algebraic variety $V_n$ topological invariants of $V_n$? If not, determine all those linear combinations of the $h^{p,q}$ and the Chern characteristic numbers which are topological invariants.
The answer to the first question is no. Some specific counterexamples are as follows. In 1986, Gang Xiao constructed two complex surfaces $S$ and $S'$ which are homeomorphic but not diffeomorphic and have different Hodge numbers. This shows that Hodge numbers are not topological invariants. But even further, one can show that $S \times S$ and $S' \times S'$ are diffeomorphic via an orientation-preserving diffeomorphism. $S \times S$ and $S' \times S'$ will still have different Hodge numbers, so this example shows that Hodge numbers are not even an invariant of oriented diffeomorphism type. See the following two references for more detail.
Xiao, Gang. An example of hyperelliptic surfaces with positive index. Northeast. Math. J. 2 (1986), no. 3, 255–257.
Campana, Frédéric. Une remarque sur les nombres de Hodge des variétés projectives complexes (Unpublished). Available here.
So the Hodge numbers are not topological invariants. But one can ask the second part of Hirzebruch's question: which linear combinations of Hodge numbers are topological invariants? This question was fully answered in quite a strong form recently by Kotschick and Schreieder.
Theorem. (Kotschick-Schreieder) The mod $m$ reduction of a $\Bbb Z$-linear combination of Hodge numbers of smooth complex projective varieties is
An oriented homeomorphism invariant or an oriented diffeomorphism invariant if and only if it is congruent mod $m$ to a linear combination of the signature, the even-degree Betti numbers and the halves of the odd-degree Betti numbers.
An unoriented homeomorphism invariant in any dimension, or an unoriented diffeomorphism invariant in complex dimensions $n \neq 2$, if and only if it is congruent mod $m$ to a linear combination of the even-degree Betti numbers and the halves of the odd-degree Betti numbers.
Their paper also classifies which linear combinations of both Hodge and Chern numbers are topological invariants. See the following for this interesting work:
Kotschick, Dieter and Schreieder, Stefan. The Hodge ring of Kähler manifolds. arXiv:1202.2676v2 [math.AG].
Now after all that, you may ask if there are some Hodge numbers which are invariant under some equivalence relation for some class of manifolds. There are some well-known results here.
Hodge numbers are homeomorphism invariants of complex curves and surfaces.
The Hodge numbers $h^{p,0}$ are birational invariants of smooth projective varieties. Not all Hodge numbers are birational invariants, as one can see by considering the blowup of a smooth projective variety.
Hodge numbers are an invariant of Kähler manifolds under deformation equivalence. Recall that two complex manifolds $X$ and $Y$ are deformation equivalent if there exists a proper, holomorphic submersion $\pi: E \longrightarrow B$ such that $X = \pi^{-1}(b)$ and $Y = \pi^{-1}(b')$ for some $b, b' \in B$. More generally, if $\pi: E \longrightarrow B$ is a proper, holomorphic submersion and $b_0 \in B$ is such that $\pi^{-1}(b_0)$ is Kähler, then there is a neighborhood $U$ of $b_0$ such that the Hodge numbers of all the $\pi^{-1}(b)$ are equal for all $b \in U$.
This is not a conclusive answer, but here's how I would approach this.
Let $\Delta_n(M)$ be the Abelian group of singular $n$-chains and $\Delta^n(M;\mathbb{R})=\mathrm{Hom}_\mathbb{Z}(\Delta_n(M),\mathbb{R})$ the $\mathbb R$-valued signular $n$-cochains. There are two subcomplexes that are relevant to the question: the complex $\Delta_*^\infty(M)$ of smooth singular chains (as explained in Bredon's book, for example), and the complex $\Delta_c^*(M)$ of compactly supported cochains, i.e. singular cochains that vanish on all chains with image outside of a compact set (which depends on the cochain).
Now, without having a reference or a proof, I would bet some money that the inclusion $\Delta_*^\infty(M)\hookrightarrow\Delta_*(M)$ is a chain homotopy equivalence. This should, in turn, dualize to chain homotopy equivalences
$$\Delta^*(M;\mathbb R)\to\Delta^*_\infty(M;\mathbb R) \quad\text{and}\quad \Delta^*_c(M;\mathbb R)\to \Delta^*_{\infty,c}(M)$$
where $\Delta^n_\infty = \mathrm{Hom}(\Delta_n^\infty,\mathbb R)$ and $\Delta^n_{\infty,c}$ is the compactly supported analogue.
Next, integration gives rise to chain maps
$$\Psi\colon \Omega^*(M)\to \Delta_\infty^*(M) \quad\text{and}\quad \Psi_c\colon \Omega^*_c(M)\to \Delta_{\infty,c}^*(M).$$
Bredon proves that $\Psi$ induces an isomorphism on cohomology and it should be possible to adapt the proof to show the same for $\Psi_c$.
Finally, the cohomology of $\Delta_c^*(M)$ is known a singular cohomology with compact supports, denoted by $H^*_c(M;\mathbb R)$. If $M$ has an orientation, then Poincaré duality gives isomorphisms
$$H^{n-i}_c(M;\mathbb R)\cong H_{i}(M;\mathbb R)$$
with singular homology on the right hand side. And if everything above goes through as claimed, then the left hand side is isomorphic to compactly supported de Rham cohomology $H^{n-i}_{dR,c}(M)$.
I'll try to find some references later. Other duties are calling.
Best Answer
Here is a version of the $\bar \partial$-Poincaré lemma as formulated in [1]
Now to your question:
This is not immediately possible, because if you formulate a similar $\partial$-Poincaré lemma, and apply it to a holomorphic form $\alpha$, you do not know if $\beta$ is holomorphic as well. However, it is still true that one can build a holomorphic de Rham complex, using the sheaves $\Omega^k$ of holomorphic differentials. It looks like one would expect: $$0 \to \mathbb C_X \to \mathcal O_X \xrightarrow{\partial} \Omega^1_X \xrightarrow{\partial} \Omega^2_X \to \dotsc \xrightarrow{\partial} \Omega^n_X \to 0$$ To show that this is an exact complex of sheaves, one can for each $k$ look at the resolution $$0 \to \Omega^k \xrightarrow{\bar \partial} \mathcal A^{k,0} \xrightarrow{\bar \partial} \mathcal A^{k,1} \dotsc$$ and those are exact by the $\bar \partial$-Poincaré lemma. Here $\mathcal A^{p,q}$ is the sheaf of $C^\infty$-forms of type $(p,q)$. Together they form a double complex $$(A^{\bullet, \bullet}, \partial, \bar \partial),$$ whose total complex is the de Rham complex $(\mathcal A^\bullet, d)$ with complex coefficients, as $\mathcal A^k = \bigoplus_{p+q=k} \mathcal A^{p,q}$. All of this shows that the holomorphic de Rham complex $(\Omega^\bullet, \partial)$ is quasi-isomorphic to the complexified de Rham complex. Hence it computes the de Rham cohomology $$H^k_{dRh}(X) \otimes \mathbb C = H^k(X, \mathbb C_X),$$ where on the right hand side I mean sheaf cohomology.
[1] Voisin, Hodge Theory and Complex Algebraic Geometry, I