There is no fundamental difference between the diagonal terms and the cross terms, you just compute a different function of the measurement outcomes. However, you need more measurements.
Let's consider a two-mode state for simplicity. The covariance matrix is:
$$
\sigma = \begin{bmatrix}
\Delta \hat{q}_a^2 & \Delta \hat{q}_a\hat{p}_a &
\Delta \hat{q}_a\hat{q}_b & \Delta \hat{q}_a\hat{p}_b \\
\Delta \hat{p}_a\hat{q}_a & \Delta \hat{p}_a^2&
\Delta \hat{p}_a\hat{q}_b & \Delta \hat{p}_a\hat{p}_b \\
\Delta \hat{q}_b\hat{q}_a & \Delta \hat{q}_b\hat{p}_a &
\Delta \hat{q}_b^2 & \Delta \hat{q}_b\hat{p}_b
\\
\Delta \hat{p}_b\hat{q}_a & \Delta \hat{p}_b\hat{p}_a &
\Delta \hat{p}_b\hat{q}_b & \Delta \hat{p}_b^2
\end{bmatrix}
$$
Here $\Delta \hat{q}\hat{p} = \frac{1}{2}\langle \{\hat{q},\hat{p}\} \rangle - \langle \hat{q} \rangle\langle \hat{p} \rangle$.
In a homodyne measurement you're continuously measuring the quadratures and get a time trace of the eigenvalues, for example: $q_a(t)$.
As you point out, the diagonal terms in the covariance matrix given by the variances of the quadratures: $\Delta q_a^2 = \langle \hat{q}_a^2 \rangle - \langle \hat{q}_a \rangle^2$ can be found by simply computing the variance of the recorded time trace $q_a(t)$.
The two-mode off-diagonal terms in the covariance matrix quantify the correlations between the modes $a$ and $b$, and these can also be computed from local measurements of the quadratures, but in this case two different parallel measurements on two modes, instead of a single measurement on one mode. For example:
$$
\Delta \hat{q}_a\hat{p}_b = \frac{1}{2}\langle \{\hat{q}_a,\hat{p}_b\} \rangle - \langle \hat{q}_a \rangle\langle \hat{p}_b \rangle =
\frac{1}{2}\langle \hat{q}_a \hat{p}_b + \hat{p}_b \hat{q}_a \rangle -\langle \hat{q}_a \rangle\langle \hat{p}_b \rangle.
$$
Because the modes $a$ and $b$ are orthogonal the quadrature operators commute, and the expression reduces to
$$
\Delta \hat{q}_a\hat{p}_b = \langle \hat{q}_a \hat{p}_b \rangle - \langle \hat{q}_a \rangle\langle \hat{p}_b \rangle.
$$
To measure this term one does two parallel homodyne measurements, one of $\hat{q}$ on mode $a$ and one of $\hat{p}$ on mode $b$. The first term in the r.h.s. is then found by taking the mean value of the product of the time-traces $q_a(t)$ and $p_b(t)$ from the two parallel homodyne measurements on the two modes.
Finally, there are terms containing two different quadrature operators of the same mode, such as $\Delta \hat{q}_a \hat{p}_a$. These can be measured by switching to a different basis. First note that:
$$
\hat{q}_a\hat{p}_a + \hat{p}_a\hat{q}_a = i(\hat{a}^{\dagger}\hat{a}^{\dagger} - \hat{a}\hat{a}).
$$
Then introduce the rotated quadratures:
$$
\hat{x}_a = \frac{e^{i\frac{\pi}{4}}\hat{a}^{\dagger} + e^{-i\frac{\pi}{4}}\hat{a}}{\sqrt{2}}, \qquad
\hat{y}_a = \frac{e^{-i\frac{\pi}{4}}\hat{a}^{\dagger} + e^{i\frac{\pi}{4}}\hat{a}}{\sqrt{2}}.
$$
It's straightforward to verify that:
$$
\hat{x}_a^2 - \hat{y}_a^2 = i(\hat{a}^{\dagger}\hat{a}^{\dagger} - \hat{a}\hat{a}) = \hat{q}_a\hat{p}_a + \hat{p}_a\hat{q}_a.
$$
The diagonal terms in the blocks on the main diagonal can therefore be found as:
$$
\Delta \hat{q}_a \hat{p}_a =
\langle \hat{q}_a \hat{p}_a + \hat{p}_a \hat{q}_a \rangle -
\langle \hat{q}_a \rangle \langle \hat{p}_a \rangle
=\langle \hat{x}_a^2 \rangle -
\langle \hat{y}_a^2 \rangle - \langle \hat{q}_a \rangle \langle \hat{p}_a \rangle,
$$
which can be evaluated by taking the variances of the time traces $x_a(t)$ and $y_a(t)$, and the means of $q_a(t)$ and $p_a(t)$.