Let's start with the case of a torus. Since you don't require $X$ to be smooth, we can reduce to this case by replacing $X$ by $X//N$. (Though actually inferring results about the nonabelian case from the abelian looks a bit hard.)
Since the DH function is piecewise continuous (even polynomial) on polyhedral chambers, we can test continuity along all straight lines. If we replace $X$ by $X//_\lambda T'$ where $T'$ is codimension $1$ in $T$, we get the DH function restricted to the line $\lambda + ({\mathfrak t}')^\perp$, which is the DH function for $S := T/T'$. So we can reduce to the case of a circle action, with moment polytope an interval in the line.
Now, the more general question is how the DH function, a piecewise polynomial of degree $\dim X - 1$, changes as one passes through the point $\lambda \in {\mathbb R}$. The answer is, there's a contribution from each component $C$ of $X^S$ with moment value $\lambda$, of a polynomial vanishing to order $(\dim X - 1) - \dim C$.
In particular, the function can only be discontinuous if $\dim C = \dim X - 1$, and in that case, $C$ must be either the source or sink of the Bialynicki-Birula decomposition for $S$. Put another way, $\lambda$ must be at an endpoint of the interval.
So for the general $T$ case, we learn that the DH function is continuous on the interior, and will only be discontinuous on a boundary facet of the moment polytope when the preimage of that facet is codimension $1$ in $X$.
EDIT: For an example, take the circle acting on $\mathbb A^3$ with weights $0,1,1$, and projectivize. The result has $DH(x)=x$ on $[0,1)$, and $DH(x)=0$ outside $[0,1]$. The fixed points are the point $[*,0,0]$ and the line $\{[0,*,*]\}$, with moment values $0,1$ respectively.