I'd like to add more information that is in line with Zhen's answer, but with slightly different hypotheses.

**Proposition:** If $C$ is cocomplete and a monad $T$ on $C$ preserves reflexive coequalizers, then the category of algebras $C^T$ is cocomplete.

Indeed, the forgetful functor $U: C^T \to C$ preserves and reflects any class of colimits that $T$ preserves, so that if $T$ preserves reflexive coequalizers and $C$ has them, then so will $C^T$. As Zhen said, we can get general coequalizers in $C^T$ if $C^T$ has binary coproducts and reflexive coequalizers, but it turns out that this follows from $C^T$ having reflexive coequalizers and $C$ having binary coproducts. See the arguments presented here for details. See particularly theorem 1, and the second corollary below it.

On the off-chance that your monad is finitary (preserves filtered colimits), this might come in handy:

**Proposition:** If $C$ is complete and cocomplete and $T$ is a finitary monad, then $C^T$ has coequalizers (and therefore is also complete and cocomplete).

See Barr and Wells, *Toposes, Theories, and Triples*, p. 267 (theorem 3.9) for a somewhat sharper statement.

For example, if products in $C$ distribute over colimits (as they do in the category of bounded posets), and your monad came from a finitary Lawvere theory $T$, this proposition would apply. See also this MO answer and this page from the ncatlab written in support of that answer.