No, the work done will not be the same, it will be greater. Some of that work will go into providing the object with kinetic energy. How do you propose to bring it to rest? If it is by applying a force in the opposite direction then the sum of TWO line integrals will result in $mgh$ as you had before.

$$W=\int\limits_{\ell=0}^{\ell=h}\vec{F}_{\rm lift}\cdot\vec{d\ell} + \int \vec{F}_{\rm stop}\cdot \vec{d\ell} =mgh,$$
where $\vec{F}_{\rm stop}$ is your stopping force and will be in the opposite direction to $\vec{F}_{\rm lift}$.