They are indeed mentioned, and in fact play a fundamental role - however, they're generally introduced after the completeness theorem is proved, which means they aren't considered separately. The corresponding algebra and its various relatives (see below) are called Lindenbaum(-Tarski) algebras.
Pedagogically, this is a slightly awkward situation: for students who haven't seen Boolean algebras before, introducing the $\mathcal{B}_?$s before we can do something with them is probably not helpful, but for students who have seen Boolean algebras before it could be quite illuminating. Most classes err towards the former population.
In the rest of this answer I'll make some comments to flesh out the picture.
First, note that in general we go beyond the picture above. Specifically, we generalize from sentences to formulas with free variables from some prescribed set $\{x_1,...,x_n\}$, and work modulo a theory $T$ (so the equivalence relation is "$T$-provable equivalence"). Call this "$\mathcal{B}_n(T)$." When $T$ is complete, the ultrafilters on $\mathcal{B}_n(T)$ are just $n$-types, and give us the type space. This is where these algebras really take off in classical model theory, e.g. in the omitting types theorem. Accordingly, this is usually the point at which this notion appears in introductory logic curricula.
Second, we can look at logics beyond first-order logic (their study being abstract model theory). While not visible in the first-order situation, there's an interesting subtlety between the completeness properties for individual sentences and sets of sentences which is mediated by the compactness of the semantics involved.
Specifically, consider the following four facts:
$\mathcal{B}_\vdash=\mathcal{B}_\models$ (that is, $\varphi\vdash\psi\iff\varphi\models\psi$ for all sentences $\varphi,\psi$).
Ultrafilters on $\mathcal{B}_\models$ correspond exactly to theories of structures.
$\vdash$ and $\models$ coincide on sets of sentences: $\Phi\vdash\Psi\iff\Phi\models\Psi$ for $\Phi,\Psi\subseteq Sent$.
Ultrafilters on $\mathcal{B}_\vdash$ correspond exactly to theories of structures.
Point (2) is just the compactness theorem; note that we can prove compactness without proving completeness first (e.g. via ultraproducts), so there is a meaningful distinction between the two. Meanwhile point (3) is the completeness theorem and point (4) follows from points (2) and (3). Moreover, these relationships are very general: they apply to abstract logics in general. (Here by "abstract logic" I mean a triple $\mathcal{L}$ consisting of a set $Sent_\mathcal{L}$ of sentences, a proves relation $\vdash_\mathcal{L}$, and a satisfies relation $\models_\mathcal{L}$ satisfying some very mild properties - not including completeness.)
Point (1), interestingly, is surprisingly weak in general! Take a non-compact entailment relation $\models_*$ on some set of sentences (say, the usual semantics for second-order logic) and let $\vdash_*$ be its "finitization" $$\Gamma\vdash_*\varphi\quad\iff\quad\exists\Gamma_0\subseteq_{fin}\Gamma(\Gamma_0\models_*\varphi).$$ Point (1) holds but point (3) fails for the resulting system. Meanwhile, point (2) of course fails since $\models_*$ is noncompact. So really what we have is that (1) by itself isn't very strong, but (1)+(2) is.
Finally, changing stances entirely note that we can choose to take the deductive structure as primitive, and ignore semantics entirely. This takes us into the realm of algebraic logic. Roughly speaking, there we consider deductive systems which are simply free algebras $A$ (thought of as the set of wffs given by some basic syntactic rules) equipped with a deduction relation $\vdash$ satisfying some very mild properties.
We can recast a deduction system $\mathfrak{D}=(A,\vdash)$ as a Lindenbaum-like algebra given by elements of $A$ preordered by $\vdash$, and taking the obvious quotient yields a partial order. However, this partial order does not in general faithfully reflect the structure of $\mathfrak{D}$! Specifically, there may be syntactic operations which are not well-defined on $\vdash$-equivalence classes. Really what we want to look at are congruences on the algebra $A$ which are related to the deduction relation $\vdash$. The book Algebraizable Logics by Blok/Pigozzi is a wonderful treatment of this topic with lots of examples, and the introduction already gives a good taste of the topic and is freely available online from the AMS.
Your analysis of $\vDash$ is fine, but you've shuffled some (important) parts of the definition of $\vDash^*$. The latter definition says: For every $\mathcal A$, if all assignments in $\mathcal A$ make all the formulas in $\Sigma$ true then all assignments in $\mathcal A$ make $\phi$ true.
The crucial difference is that here the "all assignments" quantifier is applied separately to the "make all of $\Sigma$ true" assumption and the "make $\phi$ true" conclusion, whereas in $\vDash$ the "all assignments" quantifier is applied to the whole implication "if it makes $\Sigma$ true then it makes $\phi$ true."
The underlying general fact here is that quantifiers $\forall x$ cannot be distributed across implications. $\forall x\,(P(x)\to Q(x))$ is not equivalent to $(\forall x\,P(x))\to(\forall x\,Q(x))$. Example: It's true that "if all people are American then all people are left-handed" because the antecedent and consequent are both false. But it's not true that "all Americans are left-handed."
Best Answer
Let's fix some theory $T$ in some language $\mathcal{L}$. If you wish to just consider structures you can take $T$ to be the empty theory (i.e. no axioms). We define two formulas $\phi(x)$ and $\psi(x)$ to be logically equivalent (modulo $T$) if $$ T \models \forall x(\phi(x) \leftrightarrow \psi(x)). $$ This is indeed equivalent to the definition you mention for the empty theory, and for non-empty theories it just restricts us to the models of $T$. Note that in the above definition $\phi$ and $\psi$ do not need to actually mention the variables $x$.
Your main question then was the following:
The answer is yes. Take any sentence $\psi$ that is not a tautology or contradiction. Then $\psi \wedge x = x$ is logically equivalent and has free variable $x$.
One example would be to take $\psi$ to be the sentence "there are at least 10 elements" and $T$ to be the empty theory. Then there will be structures (models of $T$) where $\psi$ is true and there will also be structures where $\psi$ is false, so $\psi$ is neither a tautology nor contradiction.
If every sentence is a tautology or contradiction (modulo $T$), so if $T$ is complete, then this can obviously not happen. Simply because there are no sentences that are not a tautology or contradiction.
Your original motivation was the following question:
You mentioned that if $\phi(x)$ is a tautology or contradiction then this cannot happen. Something more general is actually true: if $\phi(x)$ is logically equivalent to a sentence then this cannot happen. Because in such a case we have $C \models \phi(a)$ if and only if $C \models \psi$ if and only if $C \models \phi(b)$, making use of the fact that $C \models \psi$ does not depend on $a$ or $b$. This does not depend on the fact whether or not $\psi$ is a tautology or contradiction.
Finding structures $A$ and $B$ with elements $a \in A$ and $b \in B$ such that $A \models \phi(a)$ and $B \not \models \phi(b)$ heavily depends on the theory $T$ and $\phi(x)$ (also when $A = B$).