Wigner's classification of the particle representations is important, but not the only thing needed for a (quantum) field theory. In particular, you cannot expect the fields to transform in one of Wigner's representations:
A classical field $\phi$ transforming under any group $G$ is given as a section of an $G$-equivariant vector bundle over the spacetime $\Sigma$, or, equivalently, a $G$-equivariant map $\phi : \Sigma \to V_\rho$ where $V_\rho$ is some representation space of $G$ with representation $\rho$ and $\phi(\Lambda x) = \rho(\Lambda)\phi(x)$.
The definition of the field taking values in a vector space restricts it to transform in a finite-dimensional representation, hence it cannot be one of Wigner's particles. It is important that, while fields contain the creation and annihilation operators for the particles in their mode expansion, they themselves do not transform like particles. It is the Hilbert space of a QFT that must carry the proper unitary representations, not the fields.
We need a field because it encodes the dynamics of the theory - a QFT needs a map between in and out states, given by the S-matrix, which is obtained from the field action via the path integral (or the LSZ formalism or whatever approach you are most comfortable with). The mere knowledge of the Fock spaces (via Wigner's classification) does not suffice for this.
A free theory essentially gives the in/out time evolution map as the identity - states just stay the same, they don't interact at all. In this sense, you can give a free theory just by specifying the Fock spaces. You might be interesting in axiomatic formulations of QFT, e.g. the Wightman axioms, where we explicitly start from the unitary Lorentz reps + the dynamics encoded in the quantum fields, and it is explicitly demanded that the transformation of the field as an operator on the unitary reps is exactly given by the equivariant transformation $\phi(\Lambda x) = \rho(\Lambda)\phi(x) = U(\Lambda)\phi(x)U(\Lambda)^\dagger$. Even if you give the dynamics as free/trivial, you still need the field to encode them.