No. $(\omega+1)^{\omega}/\mathcal{U}$ is not complete whenever $\mathcal{U}$ is a non-principal on $\omega$. Observe that if $(\omega+1)^{\omega}/\mathcal{U}$ is complete, then $(\omega+1)^{\omega}/\mathcal{U}$ is compact in the order topology. In fact, a linearly ordered set is compact in the order topology if and only if it is complete as a linear order.
Let $U=(\omega+1)^{\omega}/\mathcal{U}\setminus\omega$ (i.e. $U$ is the collection of all non-standard natural numbers). Then $\{U\}\cup\{\{n\}\mid n\in\omega\}$ is an open cover of $(\omega+1)^{\omega}/\mathcal{U}$ with no finite subcover.