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.