No, a closed trefoil cannot appear, because it's not algebraically slice.

In more detail, I understand that in the usual Minkowski space, a "time slice" $T$ would be homeomorphic to $\Bbb R^3$. In case there are problems with this in a more general situation, I'm reading the question so that $T$ is implicitly assumed to be a copy to $\Bbb R^3$ because by the trefoil knot one normally means a certain type of knot in $\Bbb R^3$ (or in $S^3$, but I understand that a time slice would anyway be non-compact; though my argument below would work for a copy of $S^3$ just as well).

Suppose then that this $T$ intersects the given surface $F=$(long trefoil)$\times\Bbb R$ in a closed trefoil $K$. The latter is a copy of $S^1$ embedded in $F$, so by the Jordan curve theorem (for polygonal curves) it bounds a disk $D$ in $F$. Since $T\cap F=K$, this $D$ lies entirely in one of the regions $T^+$, $T^-$ of $\Bbb R^4$ bounded by $T$ (in other words, either in the future or in the past of the observer). Topologically, the two possibilities are symmetric, so we may assume $D\subset T^+$. Since $F$ is connected, $T^+\cap F=D$.

If $T^+$ happens to be diffeomorphic to $\Bbb R^3\times [0,\infty)$ we conclude immediately that $K$ is a slice knot. But the trefoil is not slice.

The general case follows by unraveling a standard proof that the trefoil is not slice. By Mayer-Vietoris and Seifert-van Kampen $T^+$ is contractible. Also, a Seifert surface of $K$ in $T$ capped off by the disk $D$ bounds a $3$-manifold in $T^+$. [Indeed, if $G$ is an infinite Seifert surface (properly embedded in $\Bbb R^3$) for the long trefoil, some perturbation of $G\times\Bbb R$ will meet $T^+$ in a $3$-manifold $M$ with $\partial M=D\cup (M\cap T)$. Since $D$ is compact, it's easy to arrange that $M$ be compact.]
It follows that $K$ is algebraically slice, and in particular its Arf invariant vanishes. But the trefoil has nonzero Arf-invariant.