$|D^+| - |D^-|$ can be expressed in terms of the level counts.
Let $S_k$ be the set of vertices of the cube with $k$ coordinates equal to $1$. Let $c_k = |X \cap S_k|$.
Let $d_k$ be the contribution to $|D^+| - |D^-|$ from the edges between $S_k$ and $S_{k+1}$.
Weight each edge $+1$ if it is white-black, $0$ if it is between vertices of the same color, and $-1$ if it is black-white. The total weight is $d_k$. We can also compute $d_k$ by weighting each half-edge leaving $S_k$ by $0$ if it leaves a white vertex and $-1$ if it leaves a black vertex, and weight each half-edge to $S_{k+1}$ by $+1$ if it the vertex is black and $0$ if the vertex is white, so the weight of each edge is the sum of the weights of its halves. So, $d_k = (k+1)c_{k+1}- (n-k)c_k. $
$$\begin{eqnarray}|D^+| - |D^-| & = & \sum_{k=0}^{n-1} (k+1)c_{k+1} - (n-k)c_k \\\ & = & \sum_{k=0}^n (2k-n)c_k. \end{eqnarray}$$
This is negative if the sum of the coordinates of the center of mass of $X$ is less than $n/2.$
I'm not sure what you mean by the assumption that $X$ is symmetric. Since you assume $|X| = 2^{n-1},$ a reasonable possibility is that you want $X$ to be self-complementary, that $(x_1,...,x_n) \in x \iff (1-x_1,...,1-x_n) \notin X.$ If so, then it is true that $|D^+| - |D^-| \gt 0$ for $n \le 3$ by inspection, it is possible for $|D^+| - |D^-| = 0$ when $n = 4$, and $|D^+| - |D^-| \lt 0$ is possible for $n \ge 5$.
$S_4 = \lbrace (1,1,1,1) \rbrace$
$S_3 = \lbrace (1,1,1,0) \rbrace$
$S_2 = \lbrace (1,1,0,0),(1,0,1,0),(0,1,1,0) \rbrace$
$S_1 = \lbrace (1,0,0,0),(0,1,0,0),(0,0,1,0) \rbrace$
$|D^+| - |D^-| = 4(1) + 2(1) + 0(3) - 2(3) = 0.$
The analogous construction for $n \ge 5$ is $X = \lbrace (1,1,...,1) \rbrace \cup \lbrace (x_1,...,x_n,0) \rbrace \backslash \lbrace (0, ..., 0) \rbrace.$ By the above summation,
$|D^+| - |D^-| = 2n -(2^{n-1}).$
Here is a construction of a counterexample for the revised symmetry condition, that $X$ is invariant under some transitive group of symmetries acting on the coordinates. Let $n=7$ and let the symmetries by cyclic rotation.
$|S_7| = 1, S_7 = \lbrace (1,...,1) \rbrace$
$|S_6| = 7, S_6 = \langle (1,...,1,0) \rangle$
$|S_5| = 7, S_5 = \langle (1,...,1,0,0) \rangle$
$|S_4| = 7, S_4 = \langle (1,1,1,1,0,0,0) \rangle$
$|S_3| =14, S_3 = \langle (1,0,1,1,0,0,0),(1,1,0,1,0,0,0) \rangle$
$|S_2| = 21, S_2 = \langle (1,0,0,1,0,0,0),(1,0,1,0,0,0,0),(1,1,0,0,0,0,0) \rangle$
$|S_1| = 7, S_1 = \langle (1,0,0,0,0,0,0) \rangle$
By the level count formula, $|D^+|-|D^-| = 7(1)+5(7)+3(7)+1(7)-1(14)-3(21)-5(7) = -42 \lt 0.$