Short answer: yes.
Long answer:
Say that a formula is elementary if it is either an atomic formula or a formula of the form $\exists x \phi$ for some $\phi$. Examples of elementary formulas are $P(x)$, $\exists x P(x)$, $\exists x \neg P(x)$, $\exists x (\neg P(x) \vee P(x))$. On the other hand, $\neg P(x)$ and $\exists x P(x) \rightarrow Q(y)$ are not elementary. Note that every quantified sentence is an elementary formula. For the purposes of defining tautological implication, quantified sentences are treated as black boxes, with no internal structure.
(Note also that, given this definition, one could define an appropriate propositional language by just taking the elementary formulas to be the propositional variables of this language (Enderton takes this approach: cf. A Mathematical Introduction to Logic, pp. 114ff). In that case, $P(x)$, $\exists x P(x)$, $\exists x \neg P(x)$, etc., would be treated as distinct sentence letters or propositional variables. They would be the same as $P, Q, R$, just with a different shape. One could then define a formula to be a tautology if it is a tautology in the propositional language generated by taking the elementary formulas as the propositional variables. It should be clear that this approach and the approach below are basically the same.)
Define a valuation $v$ as a function from the set of all elementary formulas to $\{0, 1\}$. This valuation can easily be extended to Boolean combinations of such formulas, by taking $v(\neg \phi) = 0$ iff $v(\phi)=1)$, etc. We can then say that a first-order formula $\phi$ is a tautology if $v(\phi)=1$ for every such valuation. (These definitions were taken from Shoenfield, Mathematical Logic, p. 26, but they are pretty standard.)
The idea is that such a formula is true by its mere propositional form, no matter its first-order content. If the formula is an open formula, then the idea is that its every instance will be true merely in virtue of its propositional form, where an instance of $\phi$ is the result of (uniformly) replacing its free variables for closed terms.
Notice that your example, $\forall x E(x) \vee \forall x \neg E(x)$ is not a tautology by this definition, since, in this approach, the negation sign inside the scope of the quantifier is black boxed. (In fact, since we took the existential quantifier as primitive, this sentence would actually be an abbreviation of $\neg \exists x \neg E(x) \vee \neg \exists x \neg \neg E(x)$, but the point still stands.) So this would be treated as similar to $P \vee Q$.
Finally, this notion of tautology is rather useful, since it allows us to use resources from propositional logic when dealing with first-order logic. In particular, since tautological consequence is decidable, this may be useful in searching for proofs---if the implication between two first-order formulas reduces to a tautological implication, there is actually an algorithm to find the proof of one from the other. This is one of the ideas behind Herbrand's Theorem.
