I would propose to look at this all in the following way.

Consider states of QED containing a single electron with energy below the positron creation threshold.

The sea itself, as non observable thing, is meaningless as any non observable thing. But in case of Dirac equation, it is not that non observable; rather, the holes in it are positive-energy and positive-charge particles - positrons. So, the Dirac sea is not detached from us.

But let us consider energies smaller than the hole creation threshold. Do "negative energy" states participate in the real (or dressed) electron dynamics? The answer is yes - they are involved in the radiative corrections to zeroth-order Dirac equation solutions. Often they are called "virtual particles".

In order to simplify things, let us consider a spinless non relativistic particle in a 1D box: $0\le x\le L$ with some poitential in it, say a linear potential $U(x)=k\cdot x$. The boundary conditions for the solutions are reflecting: $\Phi(0)=0,\; \Phi(L)=0$. And the initial condition is such that only the ground state may exist - because of lack of energy to be in the excited states. In that case the exact solution is well known - it is a combination of Airy functions $\Phi_0 (x)\cdot \text{e}^{-iE_0t}$ with a certain ground energy $E_0(L,k)$. No excited state can be found experimenatlly in the ground state.

On the other hand, we may solve this well posed problem by the perturbation theory with considering the potential $U(x)$ a perturbation to the pure "box solutions" $\varphi(x)^{(0)}=\psi_n(x)\sqrt{\frac{2}{L}}\sin\left(\frac{\pi\cdot n}{L}\right), \; E_n^{(0)}\propto \left(\frac {\pi \cdot n}{L}\right)^2$, $$\Phi_0 (x,t) = \text{e}^{-i\cdot t\sum_{n\ge 0}^{\infty} E_0^{(n) }} \cdot\sum_{m\ge 0}^{\infty} C_m\varphi^{(0)}_m(x)\;\;\; (1).$$ The zeroth-order solutions $\varphi^{(0)}_n(x)$ are some sort physical as they may be decent approximations to the exact solutions in the solutions like this one: $$\Phi(x,t)=\sum_{n\ge 0}^{E_{\text{max}}} A_n \varphi_n(x)\text{e}^{-iE_n t}\approx \sum_{n\ge 0}^{E_{\text{max}}}A_n^{(0)}\varphi_n(x)^{(0)}\text{e}^{-iE_n^{(0)} t}\;\;\; (2).$$ They may be "found" experimantally.

However, if we express, for example, the exact ground state solution via zeroth-order ones (1), they may not be considered physical since they are just numerical correction to the initially inexact $\varphi_0(x)$. One can call such numerical corrections as due to "virtual states", but nevertheless they may not be found experimentally (except for the zeroth-order one, of course).

Here the excited states $E_{n\gt 0}$ may be considered as "hole states" - those with a positive and higher energies than the ground state. Now it is up to you to decide if they "play some role".