I interpret OP's question as
Why does the action $S$ carry definite Grassmann parity, and in particular, why is $S$ Grassmann-even?
This question is actually not restricted to just supersymmetric models, but could be asked for any theory with Grassmann-odd variables.
It is true that the overwhelming majority of published literature have $S$ manifestly Grassmann-even.
At the classical level ($\hbar=0$), there is in principle nothing wrong with an action $S$ with indefinite or odd Grassmann parity, as long as the corresponding Euler-Lagrange equations reproduce the classical equations of motion.
At the quantum mechanical level, the situation changes. For instance, in the path integral formalism, it would be rather weird to consider a Boltzmann-factor $\exp(\frac{i}{\hbar}S)$ where the action $S$ has indefinite or odd Grassmann parity. (If one would like to consider such a construction, one would have to insert appropriate projection operators to ensure that the overlaps and amplitudes remain Grassmann-even.) Recall that no measuring device in an experiment is going to measure a Grassmann-odd number. A measuring device can only produce real outputs $\subseteq\mathbb{R}$.