Legendrian knot in 3-sphere We are given a Legendrian knot, fixed up to Legendrian isotopy, in $(S^3,\xi)$    ($\xi$ is the standard contact structure). Does it necessarily bound a symplectic surface in $(B^4,\omega)$ (again $\omega$ is the standard symplectic structure)?
 A: No, there are obstructions to this.
Suppose $L$ is a Legendrian knot that bounds a symplectic surface $\Sigma$, and $T$ a (sufficiently close) transverse push-off (see Etnyre's notes or Geiges' Introduction to contact topology for references). Notice that $T$ can be chosen to be $C^\infty$-arbitrarily close to $L$. (This is clear from the construction, although I couldn't find a clean statement in a quick search.)
You can perturb $\Sigma$ along with $K$ while keeping it symplectic, so that it bounds $T$. Now take the double cover of the ball, branched along $T$: this gives a symplectic 4-manifold whose (strongly convex) boundary is the branched cover of the standard contact 3-sphere along $T$. Hence the branched double cover of $\xi$ along $T$ is (strongly) symplectically fillable.
There are plenty of transverse knots whose branched cover is overtwisted, e.g. if $T$ is transverse stabilisation (see, for example, Harvey, Kawamuro, Plamenevskaya, On transverse knots and branched cover, which makes for a very pleasant read). To be even more concrete, take the unknot with Thurston-Bennequin number -3 and rotation number 0: this cannot bound a symplectic surface in the 3-sphere.

I guess that there could be a more elementary argument along the following lines: attach a Weinstein handle along $L$, look for a symplectic approximation of the core disk (relative to its framed boundary). You obtain a symplectic manifold with $H_2=\mathbb{Z}$, whose Chern class is determined by the rotation number (as shown by Gompf), and you can apply the adjunction formula. My guess is that if $tb(L)$ is less than $2g_*(L)-1$ (where $g_*$ is the slice genus) you have no chances.

Finally, one could wonder what happens if you ask for a Lagrangian surface rather than a symplectic one: in this case tb needs to be $2g_*(L)-1$ and rot needs to be 0. There's a nice paper by Chantraine on the subject.
A: The knot must be quasipositive.  You can perturb it to be transverse as in Marco Golla's answer, and then a theorem of Boileau and Orevkov (in Quasi-positivité d'une courbe analytique dans une boule pseudo-convexe, MR1836094) says that transverse knots which bound symplectic surfaces in $B^4$ must be quasipositive.
