Here is a counterexample, which works in ZFC without any
additional large cardinal or other extra hypothesis. This
argument, which verifies the guess I made in my original answer, is the result of a conversation I had with Arthur
Apter.

The example involves the forcing $\mathbb{S}$ to add a
stationary non-reflecting subset of $\omega_2$, that is, a
stationary set $S\subset\omega_2$, such that $S\cap\gamma$
is not stationary for any $\gamma\lt\omega_2$ of cofinality
$\omega_1$. Conditions in $\mathbb{S}$ consist of bounded
sets $s\subset\omega_2$ satisfying the condition for all
$\gamma\leq\sup(s)$. The forcing is
$\lt\omega_2$-strategically closed, since in the game where
players play a game of length $\omega_2$, with player II
playing at limit stages, player II may invent an imaginary
club set $c$ which is extended as play proceeds, and she
ensures that this club remains disjoint from the conditions
$s$ that are played. Thus, the forcing adds no new subsets
of $\omega_1$ and in particular, preserves $\omega_2$.
Also, it follows that the generic set $S\subset\omega_2$
added by $\mathbb{S}$ is indeed stationary. Note that
$\mathbb{S}$ has no $\leq\omega_1$-closed dense subset,
since with such a highly closed dense subset we would be
able to construct an initial segment of $S$ that contains a
club of order type $\omega_1$, which would violate the
non-reflecting property.

Next, let $\mathbb{T}$ be the forcing to destroy the
stationarity of the set $S$ added by $\mathbb{S}$, by
adding a club set $C\subset\omega_2$ with $S\cap
C=\emptyset$, using closed initial segments. A bootstrap
argument shows that the combined forcing
$\mathbb{S}\ast\mathbb{T}$ has a dense subset that consists
essentially of $(s,c)$, where $s\subset\gamma=\sup(s)$ and
$c$ is a closed set containing $\gamma=\sup(c)$ with $s\cap
c=\emptyset$. This dense set is $\leq\omega_1$-closed, and
thus the combined forcing $\mathbb{S}\ast\mathbb{T}$ is
forcing equivalent to $\text{Add}(\omega_2,1)$. So the
situation is that $\mathbb{S}$ makes the regrettable faux
pas of creating a stationary non-reflecting set, but
$\mathbb{T}$ apologizes, and the combination
$\mathbb{S}\ast\mathbb{T}$ is completely mild.

So now we can build the counterexample to Cantor-Bernstein
for forcing. Let $\mathbb{P}=\text{Add}(\omega_2,1)$, and
let $\mathbb{Q}=\mathbb{P}\ast\mathbb{S}$. Clearly
$\mathbb{P}$ is a factor of $\mathbb{Q}$, and $\mathbb{Q}$
is a factor of $\mathbb{P}$, precisely because
$\mathbb{Q}\ast\mathbb{T}=\text{Add}(\omega_2,1)\ast\mathbb{S}\ast\mathbb{T}=\text{Add}(\omega_2,1)\ast\text{Add}(\omega_2,1)\cong\text{Add}(\omega_2,1)=\mathbb{P}$.
So each embeds completely into the other, but they are not
forcing equivalent, because $\mathbb{P}$ has a
$\leq\omega_1$-closed dense subset, but $\mathbb{Q}$ does
not, and this is a property preserved by forcing
equivalence.

The argument easily generalizes to higher cardinals than
$\omega_2$ (but not for $\omega_1$).

Here is the original answer:

Here is a counterexample, but it uses a large cardinal. I
expect that we will be able to eliminate the large
cardinal, perhaps by constructing a similar example down
low.

Suppose that $\kappa$ is weakly compact. Let
$\mathbb{P}=\text{Add}(\kappa,1)$ be the forcing to add a
Cohen subset of $\kappa$ by initial segment. Let
$\mathbb{Q}=\text{Add}(\kappa,1)*\mathbb{T}$ be the forcing
that first adds a Cohen subset to $\kappa$, and then forces
to create a $\kappa$-Suslin tree.

Clearly, $\mathbb{P}$ is explicitly a forcing factor of
$\mathbb{Q}$. For the converse direction, observe that the
forcing $\mathbb{T}$ to create the $\kappa$-Suslin tree can
be followed by the forcing that destroys this Suslin tree
$T$, by forcing $\mathbb{D}$ to cover it with $\kappa$-many
branches. The combined forcing $\mathbb{T}\ast\mathbb{D}$
is actually isomorphic to the forcing consisting of trees
of height less than $\kappa$ that are already covered by
the branches. This forcing is ${\lt}\kappa$-closed and
hence isomorphic to $\text{Add}(\kappa,1)$. It follows that
$\mathbb{Q}\ast\mathbb{D}$ is the same as
$\mathbb{P}\ast\mathbb{T}\ast\mathbb{D}$, which is the same
as $\text{Add}(\kappa,1)\ast\text{Add}(\kappa,1)$, which is
forcing equivalent to $\text{Add}(\kappa,1)$, which is
$\mathbb{P}$. Thus, we have argued that a further forcing
extension of $\mathbb{Q}$ is isomorphic to $\mathbb{P}$,
and so $\mathbb{Q}$ is a factor of $\mathbb{P}$.

But the forcing notions $\mathbb{P}$ and $\mathbb{Q}$ are
not always equivalent. For example, it is possible to make
the weak compactness of $\kappa$ indestructible by
$\text{Add}(\kappa,1)$, that is, by $\mathbb{P}$, but the
weak compactness of $\kappa$ is always destroyed by
$\mathbb{Q}$, since this adds a $\kappa$-Suslin tree, which
is incompatible with $\kappa$ being weakly compact.

This forcing was the basis of Kunen's argument that weak
compactness is not downwards absolute. In the forcing
extension where the $\kappa$-Suslin tree is created,
$\kappa$ is not weakly compact, but the weak compactness is
recovered once one destroys the tree, since the combined
forcing is just $\text{Add}(\kappa,1)$.

By making the preparatory forcing part of $\mathbb{P}$ and
$\mathbb{Q}$, one can show that whenever $\kappa$ is weakly
compact, then there are forcing notions that are factors of
each other, but not forcing equivalent.

I suspect that a similar example can be made down low at
the level of $\omega_2$, but I have to think it through.
(If someone else can do this, please post an answer.)