No, clearly not, because you could put junk on top. 

But even if you avoid this by insisting that the map is surjective, there are counterexamples. Consider the map of [Eric Wofsey's recent answer](https://mathoverflow.net/a/194954/1946), where he considered the partial order $L(X)$, where $X$ is an antichain of pairwise incomparable elements, and we add $0$ and $1$ to bound it. This is a complete lattice, and it admits an order preserving surjection to any bounded partial order on $X$. Such an order may not be a complete lattice, and so this provides numerous counterexamples.