The answer is yes. But there are several comments that need to be made:
There is only one empty set. So it is better to say that $A=B=\emptyset$ rather than saying that "both $A$ and $B$ are empty sets", as the latter erroneously suggests that there is more than one. This is because two sets are equal precisely if they have the same elements. So any two empty sets are equal, since they have precisely the same elements (namely, none).
I assume by "context" you mean that the complement of $B$ is computed with respect to the "universal set." In the standard system of set theory, there can be no "universal set", as assuming its existence leads to problems (Russell paradox). [Though, yes, it is usual to talk of a "universal set" as a way to delineate what objects we are interested in.]
The reason why $A$ is contained in the complement of $B$ is that $A$ (being the empty set) is a subset of any set. This is because we define "$A$ is contained in $C$" to mean that any element of $A$ is also an element of $C$. Now, since nothing is an element of $A$, this condition is satisfied in this case (one typically says that it is satisfied vacuously.)
Let's make a community wiki answer based on the above comments.
Given the subset
$A \subset \Omega$
The relative complement $A^c$ in $\Omega$ is defined as the set of elements of $\Omega$ that are not elements of $A$.
So in the case that $A = \Omega$, the relative complement of $A$ in $\Omega$ is equal to $\emptyset$.
That is, for every set $x$
we have $x \notin A^c$.
In particular we have $\emptyset \notin A^c$.
However, this is probably the "wrong" question to ask, because $\emptyset$ might not be an element of $\Omega$ at all.
Notice that the $\emptyset$ is a subset of every set, so in particular $\emptyset \subset \Omega$.
This is because to check whether $X \subset Y$ you take an arbitrary element of $X$ and check that it is also in $Y$. If $X$ is empty, then there is nothing to check.
Best Answer
Assuming that in your set theory it is consistent to talk about the universal set and it is $\{x:x=x\}$, then its complement is $$ \{x:x\ne x\} $$ so no set can belong to it. In other words, the complement of the universal set is the empty set.
It's true that $\emptyset$ is a subset of any set, but this has no consequence on the fact above.