Can we divide two vectors?
No, in general you cannot divide one vector by another. It is possible to prove that no vector multiplication on three dimensions will be well-behaved enough to have division as we understand it. (This depends on exactly what one means by 'well-behaved enough', but the core result here is Hurwitz's theorem.)
Regarding force, area and pressure, the most fruitful way is to say that force is area times pressure: $$ \vec F=P\cdot \vec A. $$ As it turns out, pressure is not actually a scalar but a matrix (or, more technically, a rank 2 tensor). This is because, in certain situations, an area with its normal vector pointing in the $z$ direction can also experience forces along $x$ and $y$, which are called shear stresses. In this case, the correct linear relation is that $$ \begin{pmatrix}F_x\\ F_y \\ F_z \end{pmatrix} = \begin{pmatrix}p_x & s_{xy} & s_{xz} \\ s_{yx} & p_y & s_{yz} \\ s_{zx} & s_{zy} & p_z\end{pmatrix} \begin{pmatrix}A_x\\ A_y \\ A_z \end{pmatrix}. $$ In a fluid, shear stresses are zero and the pressure is isotropic, so all the $p_j$s are equal, and therefore the pressure tensor $P$ is a scalar matrix. In a solid, on the other hand, shear stresses can occur even in static situations, so you need the full matrix. In this case, the matrix is referred to as the stress tensor of the solid.
As an aside, you can actually divide two vectors. The only question is how do you want to interpret the objects and more importantly the operation.
For example, you can map the vectors to an object in a quaternion space quite simply as:
$$ \phi:V \rightarrow H: \vec{v} \mapsto (0,\vec{v}) , $$
and then division is well defined. But your answer will be, in general, quite obviously, a general quaternion $(r,\vec{u})$, and you then need a physical interpretation for this.
In the specifics of your question, you see, the objects and the operation are fixed by nature. Force and area are vectors related by a tensor called pressure as:
$$ \vec{F} = P \vec{A}, $$
where the operation of $P$ on $\vec{A}$ is defined to be the tensor action. In this setup there is no unique way to define division of two vectors to produce a tensor: the definition of the operation admits no sensible inverse.
To define vector division as the scalar result of one vector "divided" by another, where the scalar times the denominator vector would then give us the numerator vector, we can write the following: \begin{align*} \vec u&=w\vec v\\ \vec u\cdot\vec v&=w\vec v\cdot\vec v\\ \therefore w&=\frac{\vec u\cdot\vec v}{v^2} \end{align*}
The math for a scalar quotient works. That is one way to divide out a vector