They key distinction is that it's observables, not operators, that evolve in the Heisenberg picture. This might sound like semantics at first, given that in most introductory presentations "(Hermitian) operator" is often used as a synonim of "observable". However, operators are just mathematical objects, without an intrinsic physical meaning, and they can be used in the definition of observables, states, or in other ways. Therefore, there isn't an intrinsic rule on how operators evolve in the Schrödinger or Heisenberg picture, it depends on the role they play.
In the OP, the creation operator is used to define a state, so it doesn't evolve in the Heisenberg picture. On the other hand, if the creation operator defines an observable (for example, when you expand the field observables in normal modes), then it evolves as $a^{\dagger}_p \mapsto a^{\dagger}_{p,\mathrm{H}}(t):= e^{iHt} a^{\dagger}_p e^{-iHt}$.
For completeness, we can look at the evolution of the creation operator in the Schrödinger picture, when it is used to define a state as in the OP (this was mentioned in another post, but with a small mistake). The state evolves as
\begin{equation}
a^{\dagger}_p|0\rangle \mapsto e^{-iHt} a^{\dagger}_p|0\rangle
\end{equation}
(note, this is not the same as $a^{\dagger}_p|0(t)\rangle$). We can insert $\mathbb{I} = e^{iHt} e^{-iHt}$ between the operator and the vector to get
\begin{equation}
a^{\dagger}_p|0\rangle \mapsto
a^{\dagger}_{p,\mathrm{S}}(t) |0(t)\rangle_{\mathrm{S}},
\end{equation}
where
\begin{align}
a^{\dagger}_{p,\mathrm{S}}(t) &:= e^{-iHt} a^{\dagger}_p e^{iHt}, \\
|0(t)\rangle_{\mathrm{S}} &:= e^{-iHt} |0\rangle.
\end{align}
Note that the "Schrödinger evolution" of the operator $a^{\dagger}_{p,\mathrm{S}}(t)$ is not the same as the time evolution of an observable in the Heisenberg picture. In this case, they are related by $a^{\dagger}_{p,\mathrm{H}}(t) = a^{\dagger}_{p,\mathrm{S}}(-t)$. Don't take this as a rule though, it really depends what you are using the operators for!
Finally, it's worth mentioning that, similarly to how operators are not always observables, vectors are not necessarily states, so they might evolve or not in either picture. For example, if you calculate transition amplitides $\langle \phi | \psi \rangle$, the vector $|\phi\rangle$ plays the role of an observable, so it will evolve in the Heisenberg picture and remain invariant in the Schrödinger picture (while $|\psi\rangle $ does the opposite). This is necessary to preserve the transition amplitude $\langle \phi | e^{-iHt}|\psi \rangle$ in all pictures.
If this seems confusing, remember that (as mentioned in another post), the key is to reproduce the relevant physical quantities, which are the time-evolving expectation values
\begin{equation}
\langle O(t)\rangle = \langle \psi | e^{iHt} O e^{-iHt} | \psi \rangle.
\end{equation}
Different pictures are simply obtained by attaching the evolution operators to one or another part of the expression.
