Why does the action have to be hermitian? The hermiticity of operators of observables, e.g. the Hamiltonian, in QM is usually justified by saying that the eigenvalues must be real valued. 
I know that the Lagrangian is just a Legendre transformation of the Hamiltonian and so is also hermitian.
I am wondering if there is some other reason that the Lagrangian must be hermitian in QFT, this also made me question, why is it necessary that the action be real? am I missing some other underlying principle here?
 A: The classical action (of particles or fields) has to be real, because this means a real classical Lagrangian.
This is needed because (canonical) momenta are obtained (for instance for a particle) from the Lagrangian by $p_i = \frac{\partial \mathcal {L}}{\partial \dot x^i}$ , and momenta are real.
In QM or QFT, the action has to be understood as a phase, more precisely the probability amplitude to an history $H$ to happen is (in $\hbar  = 1$ units) : 
$\psi_H = e^{iS_H}$
where $S_H$ is the classical action of the history $H$.
To have the total quantum probability amplitude, you have to sum over all possible histories : 
$\psi = \int [DH]\psi_H = \int [Dh]e^{iS_H}$
So, the action does not correspond to a measurable quantity, it is a unobservable phase, it is not an observable.  
If the action was an observable, you should be able to select a particular history among all histories, in some sense, extract a specific classical behaviour from a quantum behaviour, which is not possible.
A: In quantum theory, the S-matrix is unitary in order to preserve the concept of probabilities adding up to 100%. Unitarity is implemented via an appropriate reality constraint/condition. E.g. in the path integral formulation $$Z~=~ \int {\cal D}\phi~e^{\frac{i}{\hbar}S},$$ the action $S$ should be real-valued.
