The short answer is yes. A (sufficiently smooth) conservative, divergence free vector field is always harmonic. And the only harmonic function disappearing at infinity is zero.
To get the details, you need to have a look at Hodge theory and work with differential forms. However the gist is the following:
You can define the curl in arbitrary dimensions, as an antisymmetric matrix.
$$(\operatorname{curl} F)_{ij} := \partial_i F_j - \partial_j F_i$$
The space of antisymmetric $3\times 3$ matrices just happens to be three-dimensional, so we can identify it with $\mathbb{R}^3$.
Now any (sufficiently smooth) conservative vector field still has curl zero, by the usual argument, since still
$$(\operatorname{curl} \operatorname{grad} \phi)_{ij} = \partial_i \partial_j \phi - \partial_j \partial_i \phi = 0.$$
But then for your conservative, divergence free vector field:
$$\Delta F_i = \sum_j \partial_j \partial_j F_i \stackrel{\operatorname{curl}=0}= \sum_j \partial_j \partial_i F_j =\partial_i\sum_j \partial_j F_j \stackrel{\operatorname{div}=0}=0.$$
So each component $F_i$ is harmonic. As mentioned before, you can then show, for example using the maximum principle that $F=0$ if it vanishes at infinity.
