We may assume that $I$ is maximal with $D(I) >1$. If $I$ is maximal and $V(I)$ is not contained in a line, then $V(I)$ is just three points in a triangle and and $I$ is the ideal of those $3$ points. If $I$ is contained on a line, then it either has two points, one with non-reduced structure pushing slightly off the line $(x,y^2)(x-1,y)$ or has one double point $(x^2,xy,y^2)$.
Now in each case we can do this fairly explicitly. For $3$ points, $I^n$ is the ideal of functions vanishing of order $n$ at those $3$ points. This contains a function of degree $(3/2)n$ when $n$ is even, which is the product of powers of the lines through the points.
This is optimal, because given a polynomial $f$, which is the first line raised to the power $a$ times a polynomial of degree $d-a$, the polynomial of degree $d-a$ must intersect the two points on the first line with multiplicity $n-a$, so $d-a \geq 2(n-a)$ and if $d \leq (3/2) n$, $a \geq n/2$. Then the same is true for the multiplicity of the other $3$ lines, hence $d \geq 3n/2$.
For one point with multiplicity $1$ and one point with multiplicity $2$, we can again manage $(3/2)n$ when $n$ is even, with the polynomial $y^n x^{n/2}$. Conversely if the degree is at most $(3/2)n$ the intersection multiplicity with the line $x$ is at least $2n$ so by the same logic the polynomial contains a factor of $x^{n/2}$. The remainder of the polynomial must vanish to order $n$ at $(1,0)$ hence have degree at least $n$, so the minimum is $(3/2)n$.
For the last case, $I^n$ is a homogeneous ideal generated by everything with degree above $2n$.
So we see if $D(I^n) < (3/2)n$, then $D(I)=1$.
EDIT: I made a mistake in the second case and the answer is actually no. Suppose $V(I)$ is two points. $I$ must be the product of the ideals of the two points. Wlog they are $(0,0)$ and $(1,0)$. If the ideals of both points contain the function $y$, then by chinese remainder theorem the product contains $y$. Wlog $(0,0)$ does not contain $y$. It's maximal with this property, so a length one extension of it does contain $y$, hence is of the form $(x^k,y)$ for some $n$. Length one extensions of that have the form $(x^{k+1}, xy, y^2, ax^k+ by)$ and we must have $a \neq 0$. If $b =0$, the ideal contains $(x^k, xy, y^2)(x-y,y)$, which is contained in the ideal I described.
Otherwise by scaling $y$ and removing unnecessary equations we may write it as $(y-x^k, x^{k+1})(x-1,y)$.
Unfortunately this ideal to the power of $k$ includes $y^{k+1}$, because:
$$ y^{k+1} = (y-x^k + x^k)^{k+1} = \sum_{i=0}^{k+1} \begin{pmatrix} k+1 \\ i \end{pmatrix} (y-x^k)^i x^{k (k+1-i) } $$
Because $k (k+1)- i \geq (k+1) (k-i)$, we lose.