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.

More generally, I claim that if $\mathcal{U}$ is a non-$\sigma$-complete ultrafilter on an index set $I$, and $X_{i}$ is a totally ordered set whenever $i\in I$, then the ultraproduct
$\prod_{i\in I}X_{i}/\mathcal{U}$ is only complete if it is finite.

The ultraproduct $\prod_{i\in I}X_{i}/\mathcal{U}$ is necessarily $\aleph_{1}$-saturated. But every $\aleph_{1}$-saturated linearly ordered set is a $P$-space in the order topology. The only compact $P$-spaces are the finite spaces. Furthermore, if $X$ is an $\aleph_{1}$-saturated linearly ordered set or if $X$ is a $P$-space in the order topology, then one can easily show that no strictly increasing sequence $(x_{n})_{n\in\omega}$ in $X$ has a least upper bound, and no strictly decreasing sequence $(x_{n})_{n\in\omega}$ in $X$ has a greatest lower bound.