Huneke only states an inclusion. But you are right, one gets a slightly stronger statement, with weaker assumptions. Namely, for any ring $R$, and any sequence $x_1,\dots,x_n$ a of elements of $R$, consider the following two properties (both implied by property $2)$ of $d$-sequences) :

$(a)$ For any $j \leq n$, one has $((x_i)_{i < j} : x_j^2) = ((x_i)_{i < j} : x_j)$.

$(b)$ For any $j \leq k \leq n$, one has $((x_i)_{i < j} : x_j) \subseteq ((x_i)_{i < j} : x_k)$.

Now let $X = (x_1,\dots,x_n)$ be an ideal generated by a sequence which has properties $(a)$&$(b)$ in $R/I$, for some ideal $I$. Then one has
$$
X^m \cap I = X^{m-1}(X \cap I)
$$
for any $m \geq 1$.

**Sketch of proof:**
Induction on $n + m$.
The case $n=1$ is handled as does Huneke, using $(a)$, and the case $m =1$ is obvious. For $n,m \geq 2$, let $a$ be an element of $X^m \cap I$, which can thus be written as $a = b + x_1 c$ with $b \in (x_2,\dots,x_n)^m$ and $c \in X^{m-1}$.

Now apply the induction hypothesis to $X'=(x_2,\dots,x_n)$ and $I' = (I,x_1)$. Since $b$ is in $X'^m \cap I'$, this yields
$$
b \in X'^{m-1}(X' \cap I') \subseteq X^{m-1}(X \cap I) + x_1 X'^{m-1}.
$$
One can thus write $a = b' + x_1 c'$ with $b'$ in $X^{m-1}(X \cap I)$ and $c'$ in $X^{m-1}$. Now $c'$ is in $X \cap (I:x_1)$. By checking that Huneke's proof of $X \cap (I:x_1) \subseteq I$ in proposition $2.1$ still holds only assuming $(a)$ and $(b)$, we get $c' \in X^{m-1} \cap I$.

By induction, we get $c' \in X^{m-2}(X \cap I)$, and thus $a \in X^{m-1}(X \cap I)$.