Let $A=\emptyset$ be the empty set.
Let $B\equiv(\forall x\in A.P(x)\land \neg P(x))$
$B$ is vacuously true.
So, I conclude that vacuously true is non-Boolean because for each $x\in A$, $P(x)$ can be true and false.
Is this correct?
There are mathematical proofs based on vacuously true statement. This is why it can be relevant.
$B$ has what you mean by a Boolean (bivalent) truth value precisely because it is only vacuously true: It is true because there are no $x \in A$. And since there are no $x \in A$, there are no $x$ such that both $P(x)$ and $\neg P(x)$ is true.
Claiming something like "both () and ¬() are true" doesn't actually entirely make sense, because it is missing information, and I believe that's where the source of the confusion lies. A statement with an open variable, such as $P(x) \land \neg P(x)$, is not just true or false, it is only true or false under a particular variable assignment which maps an object from the domain to the variable $x$. We can not talk about the truth value of $P(x) \land \neg P(x)$ without talking about an object that $x$ refers to.
So whenever we say that $P(x) \land \neg P(x)$ is true, this means that we have at hand a concrete object from $A$ assigned to the variable $x$ of which $P(x) \land \neg P(x)$ is true. But there can be no such object, since $A$ is empty!
Analagously, claiming that $P(x) \land \neg P(x)$ is false would mean we are talking about a concrete object from $A$ mapped to the variable $x$ of which the statement is false. But again: Such an object can not exist, because $A$ is empty. And since there is no object in $A$ of which $P(x) \land \neg P(x)$ is is false, there is no counter witness to the universal quantifier. (Remember that the only way for a universal statement $\forall x \in A. S$ to become false is if there is a counter witness, at least one object from the domain of $A$ of which $S$ is false. If there is no counter witness, the universal statement is true.) And since there is no counter witness, $B = \forall x \in A. P(x) \land \neg P(x)$ is true. This is precisely what is called vacuous truth: $\forall x \in A. \ldots$ is true because there are no objects in $A$ at all.
But $B$ is not simultaneously false. Because that would mean we have a counter witness $x \in A$ of which $P(x) \land \neg P(x)$ is false. But as shown above -- since there are no objects to begin with, such a counter witness can not exist, and therefore $B = \forall x \in A. ...$ can not be false.
In sum, there is no interpretation, no variable assignment that makes either $B$ or the embedded statement $P(x) \land \neg P(x)$ both true and false at the same time, so the truth values are the classical Boolean truth values without gluts or gaps.
You changed your question after I had already written my full answer...
Indeed, (1) $\forall x. P(x)$ and (2) $\forall x. \neg P(x)$ are both true. They are vacuously true since $A$ is empty. But this doesn't mean that anything is true and false at the same time. The two sentences are not negations of each other: The negation is embedded deeper in the formulas. From (1) being true we may not infer that (2) is false or vice versa, because (2) does not have the shape $\neg (1)$ and (1) does not have the shape $\neg (2)$. Only if the negation is on the outside, the statements immediately have opposing truth values, but this is not the case here. We need to look deeper inside the quantified statements.
Again: Truth of open formulas is relative to variable assignments. From $\forall x \in A. P(x)$ we can conclude that $P(x)$ is true under every assignment of an object from $A$ to a varible $x$. From $\forall x \in A. P(x)$ we can conclude that $\neg P(x)$ is true and hence $P(x)$ false under every assignment of an object from $A$ to a varible $x$.
But there will be no variable assignment such that both $P(x)$ and $\neg P(x)$ are true. Why? Because there are no variable assignments to begin with -- $A$ is empty, so we never get to actually evaluating $P(x)$ and $\neg P(x)$ for any concrete value of $x$.
Again: There is no interpretation, no variable assignment under which $P(x)$ is simultaneously true and false, precisely because there are no objects in $A$ to assign to $x$ in the first place.