Yes. If B suddenly reversed direction at some point (or if we consider another frame C going in the opposite direction at the same speed and synchronized to B at the reversal point), his/her time would still appear time dilated to A. A and B would no longer be synchronized in time at their 2nd rendezvous.
The usual argument goes like this: Say A sees B reversing direction at time $t_A^0$ and position $x_A^0 = v t_A^0$. B's time at that moment is $t_B^0 = \gamma(t_A^0 - \frac{v}{c^2}x_A^0) = t_A^0/\gamma$ and he observes $A$ at position $x_B(A) = -v t_B^0 = -x_A^0/\gamma$. From the point of view of A, when A and B meet again in $x_A=0$, A's time is $t_A^0 + x_A^0/v = 2t_A^0$, while B's time is $t_B^0 + |x_B(A)|/v = 2t_B^0 = 2t_A^0/\gamma$. So A sees B as time dilated by a factor of $\gamma$.
But we still have a couple of problems: After B reverses direction, does s/he also see that A is time dilated relative to him? And how did A and B loose synchronization of their origins, even though they were synchronized in the beginning?
The answer to the first problem is again yes, B does observe A as time dilated. In order to prove this we need the answer to the 2nd problem, which is that B's reversal of direction implies a change in synchronization through a shift in coordinates, despite the fact that he does not change the coordinates of his location at the moment of reversal. The Lorentz transforms between A and B after B's reversal read
$$
\begin{eqnarray} x_A &=& \gamma (x_B - vt_B) + 2vt_A^0 \\
t_A &=& \gamma (t_B - \frac{v}{c^2}x_B)
\end{eqnarray}
$$
and
$$
\begin{eqnarray} x_B &=& \gamma (x_A + vt_A) - 2\gamma v t_A^0\\
t_B &=& \gamma (t_A + \frac{v}{c^2}x_A) - 2\gamma \left(\frac{v}{c}\right)^2t_A^0
\end{eqnarray}
$$
(All that is needed to derive these is to account for a shift in coordinate origins in the usual form of the Lorentz transforms. I am skipping the details, but we can easily verify that all coordinates verify their correct values for $t_A = t_A^0$.)
What happens as a consequence is that B sees A at a different time after his direction reversal: Right before the reversal B observed A at position $x_B(A) = - x_A^0/\gamma$, corresponding to A's time $t_A = \gamma(t_B^0 + \frac{v}{c^2}x_B(A)) = t_A^0/\gamma^2$. Right after the reversal the new Lorentz transforms show that B still observes A at position $x_B(A) = - x_A^0/\gamma$, but now this corresponds to A's time $\bar{t}_A = \left(1+\left(\frac{v}{c}\right)^2\right)t_A^0$.
On the other hand, the 2nd rendezvous still takes place at time $t_B^0 + |x_B(A)|/v = 2t_B^0 = 2t_A^0/\gamma$ for B, and at $t_A^0 + x_A^0/v = 2t_A^0$ for A. So from B's point of view, s/he meets A again after a time $\Delta t_B = 2t_B^0 -t_B^0 = t_A^0/\gamma$, while A only observes a time lapse $\Delta t_A = 2t_A^0 - \left(1+\left(\frac{v}{c}\right)^2\right)t_A^0 = t_A^0/\gamma^2 = \Delta t_B/\gamma$. In other words, B sees A undergoing time dilation, as expected.
