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 <a href="http://ncatlab.org/nlab/show/colimits+in+categories+of+algebras#reflexive_coequalizers_and_cocompleteness_23">here</a> 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 <a href="http://mathoverflow.net/a/119004/2926">MO answer</a> and this <a href="http://ncatlab.org/toddtrimble/show/Towards+a+doctrine+of+operads">page</a> from the ncatlab written in support of that answer.