If $|\beta \rangle = A(t) |\alpha \rangle$ in Heisenberg picture, then doesn't $|\beta \rangle$ depend on time? We know that states are time independent in Heisenberg picture. However, if I apply an operator to a state in Heisenberg picture, 
$$|\beta\rangle= A(t)|\alpha\rangle \equiv \exp(itH) A(0)\exp(−itH)|\alpha\rangle$$
then the state $|\beta\rangle$ will be time dependent. However, states are not time dependent in Heisenberg picture. So is $|\beta\rangle$ no longer in Heisenberg picture? 
In schrodinger picture $A(t)$ would be $A$ but  $|\alpha\rangle$ now would be $\exp (-itH)|\alpha\rangle$ thus $|\beta\rangle$ would be $A\exp (-itH)|\alpha\rangle$
 A: The statement "states are time-independent in the Heisenberg picture" can be a bit misleading, for exactly the reason that you point out. States are not measurable in quantum mechanics; only inner products (or equivalently, matrix elements) are. The Schrodinger and Heisenberg pictures are just two ways of mentally grouping the terms in a matrix element with parentheses - they aren't completely well defined when thinking about bras and kets that haven't been contracted all the way down to scalars.
So the answer to your question is largely a matter of semantics. I personally would probably say that in the Heisenberg picture, the time-dependent ket $A(t) | \psi_0 \rangle$ should not be thought of as a "state", but as a piece of an "incomplete" matrix element. But it really depends on your exact definition of the word "state", and I wouldn't argue with someone who would prefer to use different words.
A: In Schrodinger picture, let $U$ be the unitary evolution operator that corresponds to suddenly hitting the particle; explicitly it would be 
$$U \sim e^{i F t_0 \hat{x}}$$
which applies a large force $F$ for a very short time $t_0$. Given a state $|\psi \rangle$, $U |\psi\rangle$ is just what that state would be right after being hit.
In Heisenberg picture, the physical meaning of $U(t')$ is hitting a particle at time $t'$. That is, if $U(t') |\psi\rangle = |\phi\rangle$ in Heisenberg picture, then in Schrodinger picture, 
$$|\phi(t) \rangle = |\psi(t) \rangle \text{ after } |\psi\rangle \text{ is hit at time } t', \quad |\psi\rangle = |\psi(0) \rangle, \quad |\phi\rangle = |\phi(0) \rangle.$$
So $|\phi \rangle$ does depend on the parameter $t'$, which indicates the time when the hit occurred. But $t'$ is just a parameter, which I could have called $Q$ or $\mathfrak{X}$ or $\mathcal{R}$. The state $|\phi \rangle$ still doesn't change in the time $t$, since it's in Heisenberg picture. (You could have instead considered $U(t) |\psi\rangle$, but this is an unnatural quantity that is not in Heisenberg picture.)
To give an everyday analogy, how much money do you have to put in your bank account in 2018 so that you'll be as rich as me in the long run, provided that I start with some amount of money, and plan to deposit more in $t'$ years? The answer clearly depends on $t'$ because that changes how much interest I get. But "the amount of money you need to deposit in 2018" clearly can't depend on time.
