Why do all fields in a QFT transform like *irreducible* representations of some group? Emphasis is on the irreducible. I get what's special about them. But is there some principle that I'm missing, that says it can only be irreducible representations? Or is it just 'more beautiful' and usually the first thing people tried? 
Whenever I'm reading about some GUT ($SU(5)$, $SO(10)$, you name it) people usually consider some irreducible rep as a candidate field. Also, the SM Lagrangian is constructed in this way. (Here, experimental evidence of course suggests it.)
 A: 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.
A: Gell-Mann's totalitarian principle provides one possible answer. If a physical system is invariant under a symmetry group $G$ then everything not forbidden by $G$-symmetry is compulsory! This means that interaction terms that treat irreducible parts of a reducible field representation differently are allowed and generically expected. This in turn means that we will instead reclassify/perceive any reducible field in terms of their irreducible constituents.   
A: This is only semantics. A reducible representation $\mathbf R$ of the symmetry group can be decomposed into a direct sum $\mathbf R_1 \oplus \cdots \oplus \mathbf R_N$ of irreducible representations. A field that transforms as $\mathbf R$ is the same thing as $N$ fields, which transform as $\mathbf R_1, \dots, \mathbf R_N$. When talking about fundamental fields, we can therefore assume that they transform as irreducible representations of the symmetry group.
A: Irreducible representations are always determined by some numbers, labeling the representation, which correspond to the eigenvalues of some observables which are invariant under the (unitary) action of the Lie group. 
If the group represents physical transformations connecting different reference frames (Lorentz, Poincare',...), these numbers are therefore viewed as observables which do not depend on the reference frame so that they define some intrinsic property of the elementary physical system one is considering.
If the group represents gauge transformations, these numbers correspond to quantities which are gauge invariant. In this sense they are physical quantities.
Finally, it turns out that in many cases (always if the Lie group is compact), generic unitary representations are constructed as direct sums of irreducible representations. This mathematical fact reflects the physical idea that physical objects are made of elementary physical objects (described by irreducible representations)
