Can we find 3 disjoint directed Hamiltonian cycles in the cube? Let $D$ be the digraph on $2^d$ vertices with $d2^d$ edges that we obtain by directing each edge of the $d$-dimensional hypercube in both directions.

Can we partition the edges of $D$ into $d$ directed Hamiltonian cycles?

An alternative formulation would be to ask whether the hypercube has a so-called directed double cover by Hamiltonian cycles.
For $d\le 2$ the statement holds but quite embarrassingly I'm stuck already at $d=3$.
 A: This is what I managed to figure out via google.  In "Decompositions into cycles I, Hamilton Decompositions" by Alspach, Bermond and Sotteau (google books link) of 1990, a regular graph $G$ with edge-set $E(G)$ has a Hamilton decomposition if either:

(i) $\text{deg}(G)=2d$ and $E(G)$ can be partitioned into $d$ Hamilton cycles, or
(ii) $\text{deg}(G)=2d+1$ and $E(G)$ can be partitioned into $d$ Hamilton cycles and a perfect matching.

Proposition 1 of that same paper claims that the $n$-cube has a Hamilton decomposition for all $n$.  They considered the undirected graph of the hypercube, but it certainly answers your question about $D$ positively when $d$ (in the notation of your question) is even.
Their proof in the even case relies on the fact that the $2m$-cube can be written as a product $C_4\times\cdots\times C_4$ (where $C_4$ is the 4-cycle) and a corollary of a theorem of Aubert and Schneider's:

If $C$ is a cycle and $G$ is a 4-regular graph which is decomposable into two Hamilton cycles, then $C\times G$ can be decomposed into three Hamilton cycles.

I haven't had a chance to think through whether their construction helps in giving you a decomposition of your graph $D$ into Hamiltonian cycles when $d$ is odd. Their construction in that case relies on another paper by Alspach, Heinrich and Liu, "Orthogonal Factorizations of Graphs" which I didn't try to read yet.
It seems that a simple construction for these decompositions for the n-cube is still unknown.  I found some discussion of $n=6,8$ in this paper of Okuda and Song.
There are also some slides of Alspach on Hamilton Decompositions which I found here.
A: It looks like the titled question (d=3) is not directly answered: I will hint at how to
show the answer is no.
At each vertex, there are two ways that the decomposition can go.  I like
to call them black or white, or clockwise and counterclockwise.  If we look
at a cube face and pick all of one type or the other, we induce a path
around that face, so not a Hamiltonian path.  Similarly, choosing three
vertices on a face with the same color/orientation leads to a non
Hamiltonian decomposition.  So each face has two vertices of
each orientation.
However, if we have a path that goes through a white then a black
then a white vertex, it will have gone tbrough exactly three vertices
of a face, and again will not yield a Hamiltonian path.  So each face
has to have two adjacent vertices of each color.
But if we have two adjacent faces sharing two vertices of the
same color, say black, then a third face adjacent to both will
have a white black white coloring which obstructs the
decomposition.  So not all faces can have the required coloring,
so there is no such decomposition into three Hamiltonian paths.
