The answer is "no" even for 3-dimensional Hadamard manifolds. Moreover $$\measuredangle[a\,^b_c]<\tfrac\pi2 \ \&\ \measuredangle[a\,^b_d]<\tfrac\pi2\ \nRightarrow\ \measuredangle[a\,^b_x]<\tfrac\pi2$$ for $x\in [c,d]$. Let me explain it in a nondirect way.

Let $M$ be a 3-dimensional Hadamard manifold; that is, a simply connected manifold with nonpositive sectional curvature. Suppose the implication holds. Consider two sets $A$ and $B$ defined by $$A=\{\,x\in M\mid \measuredangle[a\,^b_x]\leqslant \tfrac\pi2\ \text{or}\ x=a\,\}$$ $$B=\{\,x\in M\mid \measuredangle[a\,^b_x]\leqslant \tfrac\pi2\ \text{or}\ x=a\,\}$$ Since the $M$ is uniquely geodesic, both $A$ and $B$ are convex. Therefore $$S=A\cap B=\{\,x\in M\mid \measuredangle[a\,^b_x]= \tfrac\pi2\ \text{or}\ x=a\,\}$$ is a geodesic hypersurface.

On the other hand most 3-dimensional Hadamard manifolds do not have geodesic hypersurfaces; see "About every convex set..." by Alexander Lytchak and me.

The answer is "no" even for 3-dimensional Hadamard manifolds. Moreover, implication $$\measuredangle[a\,^b_c]<\tfrac\pi2 \ \&\ \measuredangle[a\,^b_d]<\tfrac\pi2\ \Rightarrow\ \measuredangle[a\,^b_x]<\tfrac\pi2$$ does not hold for any $x\in [c,d]$. Let me explain it in a nondirect way.

Choose a generic 3-dimensional Hadamard manifold $M$; that is, a simply connected manifold with nonpositive sectional curvature. Suppose the implication holds in $M$. Consider two sets $A$ and $B$ defined by $$A=\{\,x\in M\mid \measuredangle[a\,^b_x]\leqslant \tfrac\pi2\ \text{or}\ x=a\,\}$$ $$B=\{\,x\in M\mid \measuredangle[a\,^b_x]\geqslant \tfrac\pi2\ \text{or}\ x=a\,\}$$ Since the $M$ is uniquely geodesic, both $A$ and $B$ are convex. Therefore $$S=A\cap B=\{\,x\in M\mid \measuredangle[a\,^b_x]= \tfrac\pi2\ \text{or}\ x=a\,\}$$ is a geodesic hypersurface.

On the other hand generic 3-dimensional Hadamard manifolds do not have geodesic hypersurfaces; see "About every convex set..." by Alexander Lytchak and me. Therefore, we get a contradiction.