Your question appears to have nothing particular to do with quantum field theory, since the same can happen in ordinary quantum mechanics for any state that you labeled by the eigenvalue of a time-dependent observable, and in particular for time-dependent Hamiltonians.
Suppose you have a time-dependent observable $A(t)$ and an eigenstate $\lvert a_1,t_1\rangle$, which has the eigenvalue $a_1$ at time $t_1$. You'Re now wondering how it can be that we have a probability to find this state is the same as $\lvert a_2,t_2\rangle$ for $a_1\neq a_2$, but this is perfectly possible. Just let, for instance, $a_1,a_2$ be $+1/2,-1/2$ and let $A(t_1)$ be $\sigma_z$ and $A(t_2)$ be $-\sigma_z$, i.e. $A$ measures the spin in z-direction at both times, but the sign is flipped between the two states, and we have $\lvert a_1,t_1\rangle = \lvert a_2,t_2\rangle$.
Now, for the case of an interacting QFT that you ask about, you should first think about the picture-free way to write down what you're asking about: With the time-evolution operator "$U(-\infty,\infty)$" and the states $\lvert p_1,p_2\rangle$ and $\lvert k_1,k_2,\rangle$, you want to compute
$$ \langle p_1,p_2\vert U(-\infty,\infty)\vert k_1,k_2\rangle$$
and since the time-evolution is computed from a time-dependent Hamiltonian, you have no guarantee that any of the momentary eigenstates $\lvert p_1,p_2\rangle$ stays an eigenstate throughout the evolution. In fact, this cannot happen if there's a non-zero amplitude for $k_1\neq p_1$ or $k_2\neq p_2$, but there's nothing wrong with this in either picture.
Lastly, let me remark that the existence of the pictures in QFT to begin with is a contentious issue, due to Haag's theorem making the rigorous existence of a unitary equivalence between the fields acting on the asymptotically free Hilbert spaces and the interacting Hilbert spaces impossible.