Well, this still doesn't answer the atomless question, but I've got a violation of the desired implication among atomic Boolean algebras. 

Let $A$ consist of the finite or cofinite subsets of $\mathbb{N}$
that take $2k$ and $2k+1$ together, if at all. That is, $a\in A$ if
$a\subset\mathbb{N}$ is finite or cofinite and for every $k$ we
have $2k\in a\leftrightarrow 2k+1\in a$. Let $B$ be the Boolean
algebra consisting of all finite or cofinite subsets of
$\mathbb{N}$.

Note that $A$ is quasi-dense in $B$, since if $b$ is finite, then
$b$ is contained in an interval $[0,2k+1]$ for some large $k$, and
this is in $A$, and if $b$ is cofinite, then $b$ contains some
final segment interval $[2k,\infty)$, which is in $A$.

Let $C$ be the algebra generated by the elements of $B$ together
with the set $E$ of even numbers. Thus, every element of $C$ is the
union of a finite or cofinite subset of $E$ with a finite or
cofinite subset of $\mathbb{N}-E$. The algebra $B$ is dense in $C$,
since the singletons are dense, and they are finite.

Finally, $A$ is not quasi-dense in $C$, because the set $E$
contains no nonzero element of $A$, as it contains no odd numbers,
and is contained in no non-unital element of $A$, as the only
element of $A$ containing all the even numbers is the whole of
$\mathbb{N}$.

-------------------

This is my original answer, which shows merely that quasi-density is not transitive. 

The answer is no. For a counterexample, let $A$ have at least two atoms; let $B$ split one of those atoms, and let $C$ split both of them. 

More explicitly, let $A$ be the 4-element Boolean algebra with atoms $\{0,1\}$ and $\{2,3\}$. Let $B$ be the $8$-element algebra with atoms $\{0\}$, $\{1\}$, $\{2,3\}$, and let $C$ be the full power set, with atoms $\{0\}$, $\{1\}$, $\{2\}$, $\{4\}$. 

You may observe that $A$ is quasi-dense in $B$ and $B$ is quasi-dense in $C$ by inspection. But $A$ is not quasi-dense in $C$, since $\{0,2\}$ is neither above nor below any nontrivial element of $A$.