You have assumed many terms in that expansion commute where they don't. Anyway, whenever you calculate any commutator, you should always check what happens if it operates on a function (field) or state etc. That is, calculate $[\mathrm{D}_\mu , \mathrm{D}_\nu] \psi$ for a (scalar) field $\psi$. So
$[D_\mu , D_\nu] \psi = [\mathrm{D}_{\mu} \mathrm{D}_{\nu} - \mathrm{D}_{\nu} \mathrm{D}_{\mu}]\psi $
$ =(\partial_{\mu} -\mathrm{i} A_{\mu}) (\partial_{\nu} -\mathrm{i} A_{\nu}) \psi - (\partial_{\nu} -\mathrm{i} A_{\nu}) (\partial_{\mu} -\mathrm{i} A_{\mu}) \psi $
$= (\partial_{\mu} \partial_{\nu} -\mathrm{i} q \partial_{\mu} A_{\nu} -\mathrm{i} q A_{\nu} \partial_{\mu} - \mathrm{i} q A_{\mu} \partial_{\nu} - q^2A_{\mu} A_{\nu}) \psi$ $\large-$ [Same expression with $\mu$ and $\nu$ interchanged]
Then you should get
$$[\mathrm{D}_{\mu},\mathrm{D}_{\nu}]\psi=-\mathrm{i} q (\partial_{\mu} A_{\nu}-\partial_{\nu} A_{\mu}) \psi=-\mathrm{i} q F_{\mu \nu} \psi$$
or in other words
$$F_{\mu \nu}=\frac{\mathrm{i}}{q} [\mathrm{D}_{\mu},\mathrm{D}_{\nu}]$$