Consider $I = (y- x^k, x^{k+1})$.
For $k>1$ this does not contain any linear functions. It contains $xy$ so $D(I)=2$. But I claim $y^{k+1} \in I^k$, so $D(I^k) = k+1$.
By the binomial theorem
$$ y^{k+1} = (y-x^k + x^k)^{k+1} = \sum_{i=0}^{k+1} \begin{pmatrix} k+1 \\ i \end{pmatrix} \left(y-x^k\right)^i x^{k (k+1-i) } $$
In the exponent:
$$k(k+1-i) =k^2 +k - ik = (k+1)(k-i) + i$$
so this is
$$(y-x^k)^{k+1} + \sum_{i=0}^{k} \begin{pmatrix} k+1 \\ i \end{pmatrix} x^i\left(y-x^k\right)^i \left(x^{k+1}\right)^{k-i} \in I^k$$
(Boris pointed out a flaw in my earlier argument, leading me to find this counterexample.)
In general, subadditivity shows $\lim_{n \to \infty} \frac{D(I^n)}{n}$ exists, and that any fixed value of $\frac{D(I^n)}{n}$ is at least this limit. So one version of this question is about how to compare $D(I)$ to this limit. Here we show the limit can go arbitrarily close to $1$ with $D(I)=2$. By adding random linear factors, the limit can get arbitrarily close to $D(I)-1$. But probably for larger $D(I)$ the limit can be less than $D(I)$ by even more than $1$.