$\DeclareMathOperator\tX{\widetilde{X}}\DeclareMathOperator\tB{\widetilde{B}}\DeclareMathOperator\tD{\widetilde{D}}\DeclareMathOperator\Z{\mathbb{Z}}$

In fact, $B$ must intersect $D$ at least $|\text{link}(A,B)|$ times. Here, by the way, $D$ can be an arbitrary Seifert surface, not just a disk (so $A$ need not be the unknot).

The cleanest way to see this is to use covering spaces.

Let $X = \mathbb{R}^3 \setminus A$ and let $\pi\colon \tX \rightarrow X$ be the (unique) infinite cyclic cover of $X$. To calculate the linking number of $A$ and $B$, regard $B$ as a map $B\colon [0,1] \rightarrow X$ with $B(0) = B(1)$ and let $\tB\colon [0,1] \rightarrow \tX$ be a lift of $B$ to $\tX$. The points $\tB(0)$ and $\tB(1)$ differ by a deck transformation. Identifying the deck group with $\Z$, this deck transformation is an integer $n$ that equals the linking number of $A$ and $B$ (note that I'm being a little sloppy with the sign of the linking number since the above does not depend on the orientation of $A$, but this doesn't matter for your question).

The disk $D$ lifts to a bunch of disjoint surfaces $\tD_i$ indexed by $i \in \Z$. Indexing the $\tD_i$ correctly, they divide $\tX$ up into connected components $\tX_i$ such that $\tX_i$ contains $\tD_i$ and $\tD_{i+1}$.

When we lift $B$, we have to choose the starting point, and we might as well choose it to lie in $\tX_0$. The endpoint of $\tB$ will then be in $\tX_n$ where $n$ is the linking number. Assuming for concreteness that $n \geq 1$, we now come to the key point: to get from $\tX_0$ to $\tX_n$, the path $\tB$ must pass through each of $\tD_1,\ldots,\tD_n$. Looking downstairs, this means that $B$ must intersect $D$ a minimum of $n$ times, as claimed.