Property of the trace and expectation
I am not really sure which part you didn't understand. I'm assuming it's the step that goes from the rightmost expression at the top to the one at the bottom.
Assuming $\mathbf{q} = \mathbf{x} - \hat{\mathbf{x}}$ is i.i.d.
Let me give it a try...
$
\begin{aligned}
\textrm{E}[ (\mathbf{x} - \hat{\mathbf{x}})^{\mathrm{H}}  \mathbf{H}^{\mathrm{H}}\mathbf{H} (\mathbf{x} - \hat{\mathbf{x}}) ] 
&amp; = \textrm{tr} \{ \textrm{E}[ (\mathbf{x} - \hat{\mathbf{x}})^{\mathrm{H}}  \mathbf{H}^{\mathrm{H}}\mathbf{H} (\mathbf{x} - \hat{\mathbf{x}}) ] \} \\
&amp; =  \textrm{E}[ \textrm{tr} \{(\mathbf{x} - \hat{\mathbf{x}})^{\mathrm{H}}  \mathbf{H}^{\mathrm{H}}\mathbf{H} (\mathbf{x} - \hat{\mathbf{x}}) \} ]  \\
&amp; =  \textrm{E}[ \textrm{tr} \{\mathbf{H}^{\mathrm{H}}\mathbf{H} (\mathbf{x} - \hat{\mathbf{x}}) (\mathbf{x} - \hat{\mathbf{x}})^{\mathrm{H}}  \} ]  \\
&amp; =   \textrm{tr} \{\mathbf{H}^{\mathrm{H}}\mathbf{H} ~\textrm{E}[ (\mathbf{x} - \hat{\mathbf{x}}) (\mathbf{x} - \hat{\mathbf{x}})^{\mathrm{H}} ] \}   \\
&amp; =   \textrm{tr} \{\mathbf{H}^{\mathrm{H}}\mathbf{H} ~\textrm{E}[ (\mathbf{x} - \hat{\mathbf{x}}) (\mathbf{x} - \hat{\mathbf{x}})^{\mathrm{H}} ] \}   \\
&amp; =   \textrm{tr} \{\mathbf{H}^{\mathrm{H}}\mathbf{H} ~\sigma_{q}\mathbf{I}] \} \\
&amp; =   \sigma_{q} \textrm{tr} \{\mathbf{H}^{\mathrm{H}}\mathbf{H}] \}.
\end{aligned}$
where I am assuming $\sigma_{q} = \textrm{E}[  \|(\mathbf{x} - \hat{\mathbf{x}}) \|^2 ] = \frac{\textrm{E}[  \|\mathbf{q} \|^2 ]}{N_t} $.
As you can see, I am making a lot of assumptions here. I don't know if they hold since I couldn't access the paper. But I think it can shed some light on your problem (if your question is strictly about the trace operations).