Does there exists a knot $K\subset \mathbb{S}^3$ such that
In other words $K$ is not slice in $B^4$ but is slice in some rational homology ball.
Yes. The figure-eight knot is an example: it bounds a smooth slice disk in a rational homology ball. This has been proven in a bunch of different ways, going back to the 1980s. Here are a couple of relevant references:
Fintushel, Ronald; Stern, Ronald J., A (\mu)-invariant one homology 3-sphere that bounds an orientable rational ball, Four-manifold theory, Proc. AMS-IMS-SIAM Joint Summer Res. Conf., Durham/N.H. 1982, Contemp. Math. 35, 265-268 (1984). ZBL0566.57006.
Cha, Jae Choon, The structure of the rational concordance group of knots, Mem. Am. Math. Soc. 885, 95 p. (2007). ZBL1130.57034.
Akbulut, Selman; Larson, Kyle, Brieskorn spheres bounding rational balls, Proc. Am. Math. Soc. 146, No. 4, 1817-1824 (2018). ZBL1422.57081.
However, it remains unknown whether there's a knot is slice in an integer homology ball but not in $B^4$.
Rationally slice knots in $S^3$ are in abundance due to the collection of Kawauchi's theorems and the recent result of Kim and Wu:
Theorem ([1] + [2]): Any hyperbolic amphicheiral knot $K$ in $S^3$ is rationally slice.
Theorem ([3]): Any fibered, negative amphicheiral knot $K$ in $S^3$ with irreducible Alexander polynomial (called Miyazaki knot) is rationally slice.
Using these theorems, you may find plenty of rationally slice knots that are not smoothly slice.
[1]: Kawauchi, Akio. "The invertibility problem on amphicheiral excellent knots." Proceedings of the Japan Academy, Series A, Mathematical Sciences 55.10 (1979): 399-402.
[2]: Kawauchi, Akio. "Rational-slice knots via strongly negative-amphicheiral knots." Commun. Math. Res. 25.2 (2009): 177-192.
[3]: Kim, Min Hoon, and Zhongtao Wu. "On rational sliceness of Miyazaki's fibered,− amphicheiral knots." Bulletin of the London Mathematical Society 50.3 (2018): 462-476.
