The premise of the question is just false. When doing phenomenology it is useful to split a field into its irreducible components, essentially because each irrep carries its own coupling constant. But when analysing QFT from a theoretical point of view, it is convenient to consider a single "big" field in a reducible representation. So it is just not true that in QFT fields are irreducible: sometimes they are not.
For the representations that are relevant to conventional QFT, all reducible representations are completely reducible, so thinking of a single reducible rep, or a collection of individual irreps, is nothing but a matter of convenience: both descriptions carry the exact same information.
Take for example the beta function of Yang-Mills plus matter. The first coefficient is of the form
$$
b_0\sim C_2(G)-T(R)
$$
where $T(R)$ is the index of the representation for the matter fields. If $R$ is reducible, $R=R_1\oplus R_2\oplus\cdots\oplus R_n$, one has $T(R)=T(R_1)+T(R_2)+\cdots+T(R_n)$. Therefore, if there are $N_F$ copies of a certain irrep, $R=R_1^{\oplus N_F}$, one would write
$$
b_0\sim C_2(G)-N_F T(R_1)
$$
which is the formula one often finds in textbooks. Both formulas are identical, and one may or may not want to explicitly split $R$ into its irreps. The general case is the same: one can think of a single field in a rep $R$, or a collection of fields into the irreps of $R$. Both conventions are valid, and sometimes one is more useful than the other. But it is emphatically wrong to claim that all fields are irreducible.