In your particular case, it seems clear to sufficiency two conditions: the first is that $\nabla u^*(\omega) = 0,$ i.e, $u^*(\omega)$ is parallel. 

The last is quite technical so let me introduce some concepts before:

Recall the Weitzenbock formula:
$$\Delta = \nabla^*\nabla + Q^p.$$

Since $\Sigma$ is closed we can intrdocute the $L^2$ inner product of $\Omega^p(M)$ in the following way

$$(\omega,\beta) := \int_{\Sigma}\omega\wedge\star\beta = \int_{\Sigma} g(\omega,\beta)d\mu_g.$$

Therefore, $\Delta$ is self-adjoint on this inner product and $\nabla^*$ is the dual of $\nabla$ on this inner product.

In the particular case of $n=3$, $Q^p$ is nothing but the Ricci tensor on $2$-forms, i.e, if $R$ denotes the Riemann tensor, then
$$R(X,Y)\omega := \nabla_X\nabla_Y\omega - \nabla_Y\nabla_X\omega -\nabla_{[X,Y]}\omega,$$
$$\langle R(X,Y)\omega,\omega\rangle $$
the Ricci tensor corresponds to 
$$(\mathrm{Ric}~\omega)(X_1,X_2) = \sum_{s=1}^3\left(R(X_s,X_1)\omega(X_s,X_1) + R(X_s,X_2)\omega(X_s,X_2\right)),$$
where $\{X_1,X_2,X_3\}$ is an orthonormal pair.

Hence the Weitzenbock formula reduces to (for $p=2$)

$$\Delta = \nabla^*\nabla + \mathrm{Ric}.$$

Therefore, for any parallel $2$-form $\beta$ one has

$$(\Delta\beta,\beta) =\int_{\Sigma} \langle \Delta\beta,\beta\rangle =\int_{\Sigma} \langle  \nabla^*\nabla\beta,\beta\rangle  + \int_{\Sigma}\langle\mathrm{Ric}~\beta,\beta\rangle = \int_{\Sigma}|\nabla\beta|^2 +\int_{\Sigma}\langle\mathrm{Ric}~\beta,\beta\rangle = \int_{\Sigma}\langle\mathrm{Ric}~\beta,\beta\rangle.$$

Since
$$(\Delta\beta,\beta) = (d\delta\beta,\beta) + (\delta d\beta) = |\delta\beta|^2 + |d\beta|^2,$$
onde thus concludes that $u^*(\omega)$ is closed and co-closed, hence, harmonic, provided if
$$\int_{\Sigma}\langle\mathrm{Ric}~u^*(\omega),u^*(\omega)\rangle\leq 0,$$
and this is our second condition.