Let $K \subset V$ where $V$ is a standardly embedded solid torus in $S^3$. Suppose every meridian disk of $V$ intersects $K$ at least twice.
If $K$ is prime in $S^3$ and $\Sigma \subset V$ is a 2-sphere intersecting $K$ in two points, must $\Sigma$ bound a trivial arc of $K$ on the inside (the component of $S^3 – \Sigma$ lying entirely within $V$)?
If we remove the condition that every meridian disk of $V$ intersects $K$ at least twice, this isn't true – for example if some meridian disk intersects $K$ once then compressing $\partial V$ along that disk and pushing slightly to the interior of $V$ gives a 2-sphere intersecting $K$ but that has the trivial arc on the outside.
Best Answer
Suppose $K$ is prime, and consider the following picture:
$K$ is an associated $2$-tangle in the picture, and, modifying your example, instead of closing $K$ using a once-around arc, we use a twice-around arc. The green sphere is $\Sigma$, and it contains $K$ rather than the trivial arc.
We know this knot intersects every meridian disk at least twice for homological reasons.
Every such sphere $\Sigma$ corresponds to a torus $T\subset V$ by attaching the boundary of a tubular neighborhood of $K$ outside of $\Sigma$. In particular, this $T$ has the property that there exists a meridian disk inside $T$ that intersects $K$ exactly once. Conversely, any such $T$ gives a sphere by compressing along this disk.
The condition you want is that every essential torus $T$ in $V-K$ having a meridian disk inside $T$ intersecting $K$ in exactly one point is actually the boundary of a tubular neighborhood of $K$. Then every such $\Sigma$ contains a trivial arc.