The answer is yes. Moreover you don't need to assume that $B$ is nonpostively curved. (And, if you are not interested in the equality case or can afford a convex boundary, the nonpositive curvature of $A$ can be replaced by a weaker assumption that the geodesics in $A$ have no conjugate points).
This follows from a filling inequality that I have proved some time ago (and quite proud of it) and boundary rigidity results of Croke, Otal, or Pestov-Uhlmann.
There are two papers covering it:
1) S.Ivanov, On two-dimensional minimal fillings. Algebra i Analiz 13 (2001), no. 1, 26-38 (Russian); English translation in St. Petersburg Math. J., 13 (2002), no.1, 17-25. A preprint is here. In this paper the inequality $area(A)\le area(B)$ is proved under the assumptions that $A$ is free of conjugate points and has convex boundary (sorry;)) and for arbitrary $B$ such that $d_A(p,q)\le d_B(p,q)$ for all $p,q$ on the boundary.
2) S.Ivanov, Filling minimality of Finslerian 2-discs. arXiv:0910.2257, to appear in Trudy Mat. Inst. Steklov (= Proc. Steklov Inst. Math). In this paper the same thing is proved for Finslerian metrics $A$ and $B$ and without convex boundary assumptins. This is the maximum generality I can imagine. But the paper is harder to digest if you don't like Finsler metrics.
The equality of areas implies that the boundary distances of $A$ and $B$ coincide, and then you can apply boundary rigidity results mentioned below.
The case you are asking for is covered by C.Croke, Rigidity for surfaces of nonpositive curvature, Comment. Math. Helv. 65 (1990), no. 1, 150–169. He proves that if $A$ is a nonpositively curved and $B$ is arbitrary Riemannian such that $d_A(p,q)=d_B(p,q)$ for all $p,q$ on the boundary, then $A$ and $B$ are isometric. He carefully works through the details of non-convex boundaries.
I have not read J.-P. Otal's paper mentioned by Igor Rivin, but from mathscinet review it seems that he assumes that both $A$ and $B$ are strictly negatively curved.
For the same result without curvature assumptions but with convex boundary, see L.Pestov and G.Uhlmann, Two dimensional compact simple Riemannian manifolds are boundary distance rigid. Ann. of Math. (2) 161 (2005), no. 2, 1093–1110.