JHI's elegant lower bound of $8$ on $N$ is achieved by an explicit dissection.
I show my construction below; you might want to try to find
a solution yourself before proceeding $-$ it makes for a neat puzzle.
There may well be other ways to do it.
If somebody can make a "$3$-dimensional" graphic or picture of the
$8$-piece dissection, you're welcome to add it by editing my answer.
My diagrams are two-dimensional, labeling each piece with its height.
Fortunately the dissection is simple enough for this to be possible;
in particular, the eight pieces comprise four boxes and four L-shaped prisms.
This also made it possible to find the solution using just pencil and paper
on an otherwise uneventful international flight.
Begin by cutting the $6 \times 6 \times 6$ cube top to bottom
into three pieces, as shown in top view in the first square diagram.
Then cut each piece horizontally in two, dividing AB into $3+3$,$\phantom.$
C into $4+2$, and D into $5+1$. Each AB piece is then further subdivided
into a box B and an L-shaped prism A. The second diagram shows (say) the
bottom layer of four pieces, and the third diagram shows the top.
Note that the AB subdivisions are not quite the same.
(source)
Pieces with the same color will come together to form a smaller cube.
The larger C piece is a $4$-cube,
and the two A pieces form a $3$-cube as shown.
It remains to construct the $5$-cube from the remaining five pieces.
The last two diagrams show the bottom and top of the $5$-cube.
(source)
The two $5$'s are the larger B piece, rotated to span the entire
height of the cube, and the thick D piece.
The thin D piece completes the bottom, with width $1$.
The top is filled by the thinner C piece and the smaller B,
both rotated to height 4. QEF
I guess that a physical model won't make for a hard puzzle to reconstitute
into either one or three cubes (e.g. the AB, C, and D parts of the $6$-cube
are independent) but would still make a nice physical model of the identity
$3^3 + 4^3 + 5^3 = 6^3$.
This dissection is specific to the solution $(a,b,c;d)=(3,4,5;6)$ of the
Diophantine equation $a^3+b^3+c^3=d^3$; I don't know whether an $8$-piece
dissection is possible for any other solution. JHI's analysis shows that
one can never get below $8$, and in some cases even that's not possible:
if $a<b<c$ and $a+c<d$ then there's at least one corner of the $d$-cube,
say $(1,1,1)$, that contributes to the $a$-cube, but then any cell
$(x,y,z)$ with $\max(x,y,z) = a+1$ cannot connect to any corner.
This first happens for $(a,b,c;d) = (6,32,33;41)$.
What's the minimal dissection for the "taxicab" identity
$1^3 + 12^3 = 9^3 + 10^3$? JHI's corner-cutting argument shows
that at least nine pieces are needed.
Best Answer
The orientation-preserving symmetry group $G_n$ of the $n$-dimensional cube is an index two subgroup of the full symmetry group, which is $S_n\times\{\pm 1\}^n$. By the Polya-Burnside counting theorem, the number of ways to color the cube with two colors is just the average size of $2^{C}$, where $C$ is the number of cycles in the action of a random element $g$ of $G_n$ on the faces of the cube. Write $g = (\sigma, (s_i)_{i=1,...,n})$, where $\sigma$ is a permutation of the basis vectors and $(s_1,...,s_n)$ is a collection of signs such that $\prod_{i\le n} s_i$ is the sign of $\sigma$.
The strategy is to now fix the permutation $\sigma$ and average over collections of signs $(s_1, ..., s_n)$. Let $c$ be the number of cycles of $\sigma$, so that the sign of $\sigma$ is $(-1)^{n-c}$. A cycle $(i_1, ..., i_k)$ of $\sigma$ either stays a cycle of $g$ or splits into two cycles of $g$ depending on whether an odd or even number of the $s_{i_k}$'s are equal to $-1$. If we set $p_m$ to be the product of the $s_{i_k}$'s in the $m$th cycle, we see that $2^C = \prod_{m\le c} (3+p_m)$, and $\prod_{m\le c}p_m = (-1)^{n-c}$. The average of $\prod_{m\le c} (3+p_m)$ over all choices of $p_1, ..., p_c$ satisfying $\prod_{m\le c}p_m = (-1)^{n-c}$ comes out to $3^c + (-1)^{n-c}$ (easy exercise).
Since the average of $(-1)^{n-c}$ is $0$ for $n>1$ (i.e. exactly half of all permutations are odd), we just need to find the average value of $3^c$ where $c$ is the number of cycles of a random permutation $\sigma$ of $S_n$. By Polya-Burnside again, the expected size of $3^c$ is exactly the number of ways to color a set of size $n$ with $3$ colors, up to permutation of the $n$ elements, and this is just $\binom{n+3-1}{3-1} = \frac{(n+1)(n+2)}{2}$.
In general, the number of ways to paint an $n$-dimensional cube with $k$ colors comes out to
$\binom{n+\frac{k^2+k}{2}-1}{\frac{k^2+k}{2}-1}+(-1)^n\binom{n-\frac{k^2-k}{2}-1}{-\frac{k^2-k}{2}-1}.$
The second summand can be omitted if you allow orientation-reversing symmetries of the cube.