Take a complex nilpotent or solvable group $G$ with the right action by a co-compact lattice $\Gamma$ and conisder the quotient $G/\Gamma$. On this quotient right-invariant $1$-forms give a subspace of $H^{1,0}$. The group $G$ is acting on $G/\Gamma$ on the left and if it would presrve all the $1$-forms, $G$ would be abelian.
Torsten Ekedahl expained that what is following IS NOT CORRECT (the article of Hasegawa tells something different)
In fact, the simplest example of this kind is given by primary Kodaira surfaces (http://en.wikipedia.org/wiki/Kodaira_surface), they have two holomorphic $1$-forms.
These surfaces are described as quotinets of sovlable groups, for example, in an article of Keizo Hasegawa http://arxiv.org/PS_cache/math/pdf/0401/0401413v1.pdf
Let me add to Donu's mentioning Du Bois's Hodge decomposition. First of all, many feel that part of the credit is due to Deligne as Du Bois built heavily on his ideas. Then again that is probably true for many things in Hodge theory.
Anyway, Du Bois's main idea was that one can do Deligne's construction "one step earlier" in the sense that Deligne used simplicial resolutions to build his Hodge structure on the cohomology groups, and Du Bois does this in the derived category of coherent sheaves (with some conditions...) so he obtains a filtered complex (now mostly called the Deligne-Du Bois complex) which is quasi-isomorphic to the constant sheaf $\mathbb C$ and whose associated graded quotients (these are actual quotients in the category of complexes, before one passes to the derived category) are the objects $\underline{\Omega}_X^{p}$ Donu mentions.
It turns out that it follows directly from the construction that there is a Hodge-to-de-Rham (a.k.a. Frölicher) spectral sequence and from Deligne's work that it degenerates at the $E_1$ level. So, a lot of things actually work out the same way as in the smooth case if one uses $\underline{\Omega}_X^{p}$ in place of ${\Omega}_X^{p}$. For instance, the Kodaira-Akizuki-Nakano vanishing theorem also holds (as well as the existence of the Gysin map, etc) that can be used to prove a singular version of the Lefschetz hyperplane theorem cf. here.
As Donu mentioned, it is not known in general when the natural map ${\Omega}_X^{p}\to\underline{\Omega}_X^{p}$ is an isomorphism, except for $p=0$ which is the definition of Du Bois singularities. I think in general (that is, for arbitrary $p$) it is true that for toroidal singularities this is an isomorphism, or at least there is a good description of each object and the map between them.
We have a pretty good understanding of Du Bois singularities, although not complete and there are still many interesting open questions. Rational singularities are Du Bois by this, log canonical singularities are Du Bois by this. For an intro to Du Bois singularities and related stuff you can try this.
There is also an intriguing connection to singularities defined via the action of Frobenius in positive characteristic. For more on this see this and other works of Karl Schwede.
Regarding the question on having something that reflects the non-reduced structure, I
am afraid that the topological $H^i$ do not see the non-reduced structure, so I don't think you can expect any reasonable Hodge theory that remembers that.
Best Answer
Let $\Omega^{p,q}(M)$ be the $C^{\infty}$ $(p,q)$-forms. One always has a double complex with $\Omega^{p,q}(M)$ in position $(p,q)$. The cohomology in the $q$ direction is Dolbeault cohomology, the cohomology of the total complex is deRham cohomology. (In each case, essentially by definition.) Whenever you have a double complex, you get a spectral sequence. On the first page of the spectral sequence is Dolbeault cohomology; the spectral sequence converges to deRham cohomology. This is sometimes called the Hodge-de Rham spectral sequence, and sometimes called the Frolicher spectral sequence.
If $M$ is compact Kahler (in particular projective) then the spectral sequence collapses at the first page; all maps between Dolbeault groups are zero. So $H^k(M) \cong \bigoplus_{p+q=k} H^{p,q}(M)$ in this case.
If $M$ is Stein (in particular, affine) then the only nonzero Dolbeault groups are the $H^{p,0}(M)$, corresponding to holmorphic $p$-forms. The next page takes the cohomology of the complex of holomorphic $p$-forms; I'll term this ``holomorphic deRham". After that, there are no further maps, so holomorphic deRham equals deRham.
In general, the spectral sequence can be arbitrarily nondegenerate.