Let's look at bicubic planar graphs. As you say, there must be $2n$ vertices, $3n$ edges, and $n+2$ faces for some $n$. Suppose there are $F_j$ faces of length $j$; then we want lower and upper bounds on $F_4$ in terms of $n$.
Note that every 4-cycle in the graph must be a face in the planar embedding, because any chord in the 4-cycle would give you cycles of length 3, impossible in a bipartite graph. The graph you have at the end of the question is not a counterexample because it is not cubic (in a 3-regular graph, there would be two triangles on the bottom.)
Upper bound
The upper bound is $n+2$. This is obviously the maximum possible because it is the total number of faces, and it is achieved by the cube graph, where $n = 4$.
This is the only graph that achieves the bound, because it is the only cubic planar graph consisting only of squares. If all the faces are squares, then 4 times the number of faces must be 3 times the number of vertices, or
$$ 4(n+2) = 6n $$
so $n = 4$.
If we exclude this single graph, then the upper bound is $n$, and graphs with $F_4 = n$ can be found for arbitrarily large $n$ (see the prism graph at the end of this answer.)
To prove this, we need to show that we cannot have $n + 1$ 4-sided faces and a single $j$-sided face, with $j > 4$.
In this case
$$
6n = 4F_4 + jF_j = 4(n+1) + j
$$
so $j = 2n - 4$. The same logic as this answer by Henning Makholm applies:
Now start by drawing the face with $2n - 4$ sides. That uses up $2n -4$ nodes. Since the graph is cubic, each of these nodes have an extra neighbor. But if those neighbors are all different, then there would be at least $4n - 8$ nodes in the graph which is too many.
Therefore, there must be a pair of corners of the $j$-gon that share a neighbor outside the $j$-gon. Since [all other faces are 4-cycles], this is only possible if these two nodes are separated by one corner in the $j$-gon. But then it's impossible for the extra leg of the corner between them to connect to anything, which is absurd. So the graph cannot exist.
(It's not that absurd to have an edge going to another graph component
contained within a face, but such an edge would be a cut-edge, which cannot exist in a regular bipartite graph.)
Lower bound
The lower bound is an arbitrarily small multiple of $n$. We can see this by applying a bitruncation operation, which replaces every vertex by a hexagon and leaves a smaller version of each face (with the same number of sides). Thus, starting from the cube, we get a bitruncated cube (aka truncated octahedron) with 6 squares and 8 hexagons ($n = 12$).
Bitruncating that gives a polyhedron with 6 squares and 32 hexagons ($n = 36$).
Bitruncating that gives a polyhedron with 6 squares and 104 hexagons ($n = 108$).
Here are the corresponding planar graphs:
In general, bitruncating yields a graph with 3 times as many vertices and the same number of squares, so we can keep going and get $F_4$ to be less than $\epsilon n$ for any $\epsilon > 0$.
Another option would be a chamfering operation, which replaces every edge by a hexagon and every face by a smaller face of the same number of sides. Starting from the cube, this gives you 4 times as many vertices with each step, so $n$ grows even faster, while always keeping 6 squares.
Absolute bounds
As you mention, any 3-regular planar bipartite graph must have at least 6 squares (assuming you forbid digons, that is, two edges between the same vertices). One way to show this is that 3 times the number of vertices equals the sum of each face's length, so
$$\begin{align}
6n &= 4F_4 + 6F_6 + 8F_8 + \ldots \\
&\geq 4F_4 + 6(n+2-F_4) \\
&= 6n + 12 - 2F_4 \\
2F_4 &\geq 12 \\
F_4 &\geq 6.
\end{align}$$
(If you allow digons, you can get this guy with two squares: 
)
There is no absolute upper bound on $F_4$ (independent of $n$), since you can have prism graphs with arbitrarily many squares:
These all have $F_4 = n$ and are bipartite for any even $n$.
The proof is not valid. In particular, to add a new vertex, it doesn't just need to join two existing vertices, but ones that are adjacent around the polygon built so far; there is no guarantee that such a vertex exists.
The statement is not always true, either. The smallest counterexample is the Goldner–Harary graph shown below:
In this maximal planar graph, there simply is no cycle of length $n$ that visits every vertex once. A quick way to see that is to look at what happens when we delete the $5$ vertices marked with an $X$ below:
The result is $6$ isolated vertices. However, if you have an $11$-gon and delete any $5$ of its vertices, you'll leave the remaining $6$ vertices in $5$ or fewer pieces; there will be some edges between them left. So this graph cannot have an $11$-gon inside it. (In other words, the Goldner-Harary graph is not Hamiltonian because it's not $1$-tough.)
Best Answer
Yes, they are (and well spotted).
Here's an old proof I have typed up. I think you can shorten it substantially however! To do so, I'd directly consider the walk bounding a face. It consists of alternating black / white vertices. Try argue using maximality that the walk is either a 4-cycle, or a star.
Lemma 1: Every face of a 2-connected plane graph is bounded by a cycle. This is well known (see, for example, Diestel)
Theorem 1: Let $G$ be an mpb graph of minimum degree $\delta$ embedded in the plane. Then every face of $G$ is bounded by a 4-cycle if and only if $\delta \geq 2$.
Proof: Let $G = (V,E)$ be an mpb graph and consider an embedding of $G$ in the plane. If every face of $G$ is bounded by a 4-cycle, then clearly $\delta \geq 2$. Conversely, assume $\delta \geq 2$. As $G$ is bipartite, $V$ can be partitioned into two sets, $B$ and $W$, such that every edge of $G$ is incident with one vertex of $B$ and one vertex of $W$.
First, we show that $G$ is 2-connected: Assume to the contrary, and without loss of generality, that $b_1\in B$ is a cut-vertex. Let $w_1, w_2$ be vertices of different components of $G-b_1$ such that $w_1, b_1, w_2$ are consecutive vertices of the walk bounding some face $f$ of $G$. As $w_1, w_2$ are adjacent to $b_1\in B$, it must be that $w_1, w_2 \in W$. Since the minimum degree of $G$ is at least 2, $w_2$ has a neighbour $b_2\in B$ such that $w_2, b_2$ are consecutive vertices of the walk bounding $f$. Since $b_2$ and $w_1$ are in different parts of the bipartition of $G$, and are both on the boundary of $f$, the maximality of $G$ implies that $w_1b_2 \in E$. Thus, $w_1, w_2$ lie in the same component of $G-b_1$, a contradiction. Hence $G$ is 2-connected.
We now show that every face of $G$ is a 4-cycle: Since $G$ is 2-connected and bipartite, by Lemma 1, every face is bounded by an even cycle. Thus it suffices to prove that every face of $G$ is bounded by a cycle of length less than 6. Assume, to the contrary, that some face $f$ of $G$ is bounded a cycle $C$ of length $k\geq 6$ and, without loss of generality, that $f$ is in the interior of $C$. Let $b_1, w_1, b_2, w_2, b_3, w_3$ be six consecutive vertices of $C$ with $b_1, b_2, b_3 \in B$ and $w_1, w_2, w_3 \in W$. As $b_1$ and $w_2$ lie on the same face and are in different parts of the bipartition of $G$, the maximality of $G$ implies that $b_1w_2 \in E$. Since the interior of $C$ is a face, $b_1w_2$ must lie in the exterior of $C$. Similarly to $b_1w_2$, the maximality of $G$ necessitates that $w_1b_3\in E$. If $w_1b_3$ lies inside $C$, it subdivides $f$, contradicting that $f$ is a face. If $w_1b_3$ lies outside $C$, it crosses $b_1w_2$, contradicting the planarity of $G$.
Theorem 2: If $G$ is an mpb graph of minimum degree one, then $G$ is a star.
Proof: Let $G = (V,E)$ be an mpb graph with partite sets $B$, $W$, and consider an embedding of $G$ in the plane. Without loss of generality, $v\in B$ is the neighbour of a vertex of degree 1. We claim that every neighbour of $v$ has degree 1. Assume to the contrary that $v$ has a neighbour of degree greater than 1. We can find neighbours $u,w$ of $v$ with $\text{deg}(u) = 1$ and $\text{deg}(w) \geq 2$ such that $u,v,w$ lie on the same face, $f$, of $G$. Let $b\in B$ be a neighbour of $w$, other than $v$, which lies on the boundary of $f$. Since $u$ and $b$ lie on the boundary of the same face, in different parts of the bipartition of $G$, they are adjacent by the maximality of $G$, contradicting the fact that the degree of $u$ is 1.