On the Euclidean action for QCD The Euclidean action for QCD reads, (see e.g., Eq. (45) in "ABC of instantons" by Novikov, Shifman, Vainshtein, and Zakharov)
$$S_E=\int d^4 x\left[\frac{1}{4}G^a_{\mu\nu}G^a_{\mu\nu}+\psi^\dagger(-i\gamma_\mu D_\mu-im)\psi\right].\tag{45}$$
Here $\gamma_\mu$ are gamma matrices in Euclidean space, i.e., $\{\gamma_\mu,\gamma_\nu\}=2\delta_{\mu\nu}$. Sometimes, people also use the notation $\bar\psi$ to replace the above $\psi^\dagger$.
Now, we know that the operator $-i\gamma_\mu D_\mu$ is hermitian but $-im$ is not (actually it is anti-hermitian). Then the fluctuation operator as a whole, $-i\gamma_\mu D_\mu -i m$ is not hermitian. Won't this be a problem? For instance, the eigenvalues for this operator are in general complex.
How do people treat this in,for instance, lattice QCD?
 A: I think you might be missing something due to the notation used in that article. In Lorentzian signature the quantity $\psi^\dagger \gamma_0 \psi$ is a scalar. Going to Euclidean signature with $\gamma_0 = \gamma_4$, we have that $\psi^\dagger \gamma_4 \psi$ is again a scalar. However, note that the paper defines a new quantity see Equation (44) to distinguish the Euclidean space quantities from Lorentzian quantities:
$$ \hat{\bar\psi} = i \bar \psi = i \psi^\dagger \gamma_4 \qquad \hat \psi = \psi $$
Hence, 
$$ \bar \psi \psi = -i \hat{\bar \psi} \hat \psi   \implies (-i) \times \hat{\bar \psi} \hat \psi  = \bar \psi \psi \quad \text{ is hermitian}$$
However, the easiest answer the question is that in Equation (45) they establish the relationship between the Euclidean and the Lorentzian Lagrangian. Since the Lorentzian Lagrangian is hermitian so must be the Euclidean one since by definition $\mathcal{L}_E$ is constructed from $\mathcal{L}_L$ by sprinkling $i$'s so that schematically
$$ \mathcal{L}_E[\phi_E] = \mathcal{L}_L[\phi_L]$$
