There cannot be a "free coalgebra" functor, at least in what I think is the standard usage. Namely, suppose that "orange" is a type of algebraic object, for which there is a natural "forgetful" functor from "orange" objects to "blue" objects. Then the "free orange" functor from Blue to Orange is the left adjoint, if it exists, to the forgetful map from Orange to Blue.
Suppose that the forgetful map from coalgebras to vector spaces had a left adjoint; then it would itself be a right adjoint, and so would preserve products. Now the product in the category of coalgebras is something huge — think about the coproduct in the category of algebras, which is some sort of free product — and it's clear that the forgetful map does not preserve products.
On the other hand, the coproduct in the category of coalgebras is given by the direct sum of underlying vector spaces, and so the forgetful map does preserve coproducts. This suggests that it may itself be a left adjoint; i.e. it may have a right adjoint from vector spaces to coalgebras, which should be called the "cofree coalgebra" on a vector space.
Let me assume axiom of choice, so that I can present the construction in terms of a basis. Then I believe that the cofree coalgebra on the vector space with basis $L$ (for "letters") is the graded vector space whose basis consists of all words in $L$, with the comultiplication given by $\Delta(w) = \sum_{a,b| ab = w} a \otimes b$, where $a,b,w$ are words in $L$. I.e. the cofree coalgebra has the same underlying vector space as the free algebra, with the dual multiplication. It's clear that for finite-dimensional vector spaces, the cofree coalgebra on a vector space is (canonically isomorphic to) the graded dual of the free algebra on the dual vector space. Anyway, this is clearly a coalgebra, and the map to the vector space is zero on all words that are not singletons and identity on the singletons. I haven't checked the universal property, though.