No, the eigenbasis is not unstable as the Hamiltonian approaches the time of the level crossing, and this can be seen by considering an adiabatic approximation which decouples the two crossing energy levels from the rest$\def\ket#1{|#1\rangle}\def\bra#1{\langle#1|}.$
How one could be fooled into thinking the eigenbasis is unstable
First, consider why one might think that the eigenvectors are inherently unstable. Let $\ket{E_0}, \ket{E_1}$ be the $E_0$-eigenvector and $E_1$-eigenvector respectively, and let $R: \mathbb C^2 \to \mathcal H$ be an operator mapping $\ket0 \mapsto \ket{E_0}$ and $\ket{1} \mapsto \ket{E_1}$ respectively. The rate of change of $R$, one might imagine, relates directly to how quickly the eigenstates change:
$$ \ket{\dot E_j} = \tfrac{\mathrm d}{\mathrm dt} \Bigl[ R \ket{j} \Bigr] = \dot R \ket{j} .$$
Note that because $R^\dagger R = 1$ by construction, we have
$$ 0 = \tfrac{\mathrm d}{\mathrm d t} \Bigl[ R^\dagger R \Bigr] = \dot R^\dagger R + R^\dagger \dot R ,$$
which implies that $R^\dagger \dot R$ is anti-hermitian. To get at the rate of change of $R$, consider the fact that
$$ H = R D R^\dagger $$
for $D = \mathrm{diag}(E_0,E_1)$, so that
$$ \dot H = \dot R D R^\dagger + R \dot D R^\dagger + R D \dot R^\dagger ,$$
which implies that for distinct $j,k \in \{0,1\}$,
$$\begin{aligned}[b]
\bra{E_j} \dot H \ket{E_k}
&= \bra{E_j} \dot R D \ket{k} + \bra{j} \dot D \ket{k} + \bra{j} D \dot R^\dagger \ket{E_k}
\\&=E_k \bra{j}R^\dagger \dot R \ket{k} + E_j \bra{j} \dot R^\dagger R \ket{k}
\\&=(E_k - E_j) \bra{j} R^\dagger \dot R \ket{k}
\\&=(E_k - E_j) \bra{E_j} \dot R R^\dagger \ket{E_k}.
\end{aligned}$$
This would seem to imply that if $E_j - E_k$ vanishes, then the operator norm of $\dot R R^\dagger$ (and thus of $\dot R$ itself) will increase without bound unless the cross-terms of $\dot H$ in the $\ket{E_j}$-basis also vanish.
An observation which points the way forward
The key question when considering the cross-terms for $\dot H$ in the $\ket{E_j}$-basis is: how does one determine that basis to begin with, to evaluate the cross-terms? Without being able to solve for the eigenstates for times approaching the crossing, we are left only with the time of the crossing itself — and crucially, the eigenspace there is degenerate, which means that just because we have one eigenbasis in mind, does not make it the physically sensible choice.
My original conjecture (in a previous edit of the question) didn't involve the projector $\Pi$ onto $\mathrm{span}\{\ket{E_0},\ket{E_1}\}$. But it occurred to me later that it's irrelevant whether or not $\dot H$ fails to commute with $H$ if this is because some of the other eigenstates of $H$ are not eigenstates of $\dot H$. What really matters is whether $\dot H$ fails, for the two crossing eigenvalues alone (in a manner of speaking), to commute with $H$. So what we really care about is just the subspace spanned by $\ket{E_0}$ and $\ket{E_1}$, leading to the modification of the conjecture using the projector $\Pi$. But at the time of the crossing, $\Pi H = E_0 \Pi$ by that very fact: in the subspace, it is proportional to the identity, which commutes with everything. Thus we will have $$0 = [H, \Pi \dot H] = H \Pi \dot H - \Pi \dot H H = \Pi [H, \dot H].$$ That is, the conditions of the conjecture will always hold, which should indicate that the worry is over nothing.
Adiabatic restriction: a sketch
If the other eigenvalues of $H$ are bounded away from $E_0$ and $E_1$ by a constant near to the crossing, we don't really mind the extent to which $\ket{\dot E_j}$ for $j \in \{0,1\}$ overlaps the other eigenvectors of $H$: by the analysis above we expect them to do so by a finite amount. We're only really concerned with the magnitude of $\bra{E_j} \dot H \ket{E_k}$. So we may completely restrict our attention to the effective coupling of $\ket{E_0}$ and $\ket{E_1}$ by $\dot H$, which is to say that we may in effect replace $\dot H$ with $\Delta = \Pi \dot H \Pi$.
Having done so, we now have in effect a Hermitian operator $\Delta$ on a two dimensional subspace, which obviously has two eigenvectors spanning the space. These are the two common eigenvectors $\ket{\delta_0}, \ket{\delta_1}$ of $\dot H$ and $H$ at the level crossing, and the cross-terms of $\dot H$ between these two vectors will be zero near to the level crossing.
The two vectors $\ket{\delta_0},\ket{\delta_1}$ may not be eigenvectors of $\dot H$, but they do allow us to see that $R^\dagger \dot R$ may have bounded operator norm in a neighborhood of the time $T$ of the level crossing, when restricted to $\mathrm{span} \{ \ket{E_0}, \ket{E_1} \}$, if $\ket{E_0} = \ket{\delta_0}$ and $\ket{E_1} = \ket{\delta_1}$ at time $T$. For $t$ nearby to $T$, an adiabatic argument would indicate that $\ket{E_0}$ and $\ket{E_1}$ will mostly not interact with the other energy eigenstates if the evolution is slow enough, so that we expect $\ket{\delta_0}, \ket{\delta_1}$ to nearly be eigenvectors of $H$ for $t \approx T$. There is a question of how quickly $\ket{E_j}$ converges to $\ket{\delta_j}$, which determines how quickly the cross-terms $\bra{E_j} \dot H \ket{E_k}$ vanish; however, large cross-terms would correspond to large eigenvalues of $\Delta \propto \Pi H \Pi + \text{const.}$, which should cause $R^\dagger \dot R$ to quickly converge to an operator diagonal in the $\ket{\delta_j}$-basis.
The coefficients of $R^\dagger \dot R$ for the other energy levels in the energy eigenbasis should be bounded either because of eigenvalue gaps between them, or for similar reasons if they exhibit level crossings of their own.
So no: there should be no instability of the eigenbasis.
