Why is the "canonical momentum" for the Dirac equation not defined in terms of the "gauge covariant derivative"?

The identification goes as follows:

$$ \text{Kin. Mom.}~=~ \text{Can. Mom.} ~-~\text{Charge} \times \text{Gauge Pot.} $$

$$ \updownarrow $$

$$ m\hat{v}_{\mu} ~=~ \hat{p}_{\mu} - qA_{\mu}(\hat{x})$$

$$ \updownarrow $$

$$ \frac{\hbar}{i} D_{\mu} ~=~ \frac{\hbar}{i}\partial_{\mu} - qA_{\mu}(x) $$

$$ \updownarrow $$

$$ D_{\mu} ~=~ \partial_{\mu} -\frac{i}{\hbar} qA_{\mu}(x) $$

$$ \updownarrow $$

$$ \text{Cov. Der.}~=~ \text{Par. Der.} ~-~\frac{i}{\hbar}\text{Charge} \times \text{Gauge Pot.} $$

The imaginary unit $i$ is needed, e.g. because the derivative is an anti-hermitian operator (recall the usual integration-by-part proof), while the momentum is required to be a hermitian operator in quantum mechanics.