Both the cell complex, $C$, and the dual cell complex $C'$ are refined by the first barycentric subdivision $BC$. There are maps $C \to BC$ and $C' \to BC$, sending a cell $\sigma$ to the sum of all cells of the same dimension contained in $\sigma$; these maps are both quasi-isomorphisms.
So, if you allow me to formally invert quasi-isomorphisms, I'm done.
Is the question whether there is an honest map of chain complexes between $C$ and $C'$, without subdividing?
UPDATE Here is something you can do, and something you can't do.
With $C$ and $BC$ as above, and $r : C \to BC$ the refinement map, there is a homotopy inverse $s: BC \to C$. (More precisely, $C \to BC \to C$ is the identity, and $BC \to C \to BC$ is homotopic to the identity.) Working the same trick with $r' : C' \to BC$, we get quasi-isomorphisms between $C$ and $C'$ which are homotopy inverse to each other. As you will see, however, this construction is very nongeometric and inelegant.
Construction: Let $q:BC \to Q$ be the cokernel of $C \to BC$. An easy computation checks that each $Q_i$ is free. Since $C \to BC$ is a quasi-isomorphism, $Q$ is exact. An exact complex of free $\mathbb{Z}$ modules must be isomorphic to a direct sum of complexes of the form $\cdots \to 0 \to \mathbb{Z} \to \mathbb{Z} \to 0 \to \cdots$. Choose such a decomposition of $Q$, so $Q_i = A_{i+1} \oplus A_{i}$ and the map $Q_i \to Q_{i-1}$ is the projection onto $A_{i}$.
Now, consider the map $q_i^{-1}(A_i) \to A_i$ in degree $i$. This is surjective, and $A_i$ is free, so choose a section $p^1_i$. We also define a map $p^2_i$ from the $A_{i+1}$ summand of $Q_{i}$ to $BC_i$ by $p^2_i = d p^1_{i+1} d^{-1}$. In this way, we get maps $p_i = p^1_i \oplus p^2_i: Q_{i} \to BC_i$ which give a map of chain complexes.
We note that $qp: Q \to Q$ is the identity. Therefore, $1-pq$, a map from $BC \to BC$, lands in the subcomplex $C$ and gives a section $s:BC \to C$. Proof of the claim about homotopies will be provided on request.
On the other hand, here is something you can't do: Get the quasi-isomorphism to respect the symmetries of your original space. For example, let $C$ be the chain complex of the cube, and $C'$ the chain complex of the octahedron. I claim that there is no quasi-isomorphism $C \to C'$ which commutes with the group $S_4$ of orientation preserving symmetries.
Consider what would happen in degree $0$. A vertex of the cube must be sent to some linear combination of the vertices of the octahedron. By symmetry, it must be set to
$$a (\mbox{sum of the "near" vertices}) + b (\mbox{sum of the "far" vertices})$$
for some integers $a$ and $b$. But then the map on $H_0$ is multiplication by $3(a+b)$, and cannot be $1$.
I imagine you want something stronger then my first answer, but weaker than my second. I am not sure what it it, though.
Best Answer
This is a terminology question.
I think that "weak homotopy equivalence" is mostly used for maps between topological spaces (as opposed to "weak equivalence", that is used in the much broader context of model categories).
The term "quasi-isomorphism", on the other hand, is typically used for (co)chain complexes, or (co)chain complexes equipped with extra structure.
So I would say that the main difference between those two terms is that they are used in different contexts.