An incompressible and boundary incompressible connected surface is isotopic to either (a) a vertical annulus or (b) a horizontal surface.  A vertical annulus is of the form $\alpha \times I$ where $\alpha$ is an essential simple closed curve.  A horizontal surface is of the form $\Sigma \times \{t\}$.  

Here is a sketch of the proof.  Let $F$ be the given incompressible, boundary incompressible, connected surface.  Suppose that $\beta \subset \Sigma$ is essential simple closed curve.  Let $B = \beta \times I$ be the corresponding vertical annulus.  An innermost disk/outermost bigon argument simplifies the intersection between $F$ and $B$ until it is a disjoint union of either vertical arcs or horizontal curves (ie, copies of $\beta \times \{t\}$).  

Now cut $\Sigma \times I$ along $B$ to get a handlebody with a product structure.  Repeat the above argument, replacing the vertical annulus with a sequence of vertical rectangles.  

I believe that you can find all of the tools you need for this kind of thing in Gordon's lecture notes on normal surfaces.  http://homepages.warwick.ac.uk/~masgar/Articles/Gordon/normal.pdf

There is also a proof of a similar fact by Scharlemann and Thompson in their paper "Heegaard splittings of (surface)×I are standard." 

Allowing boundary compressible surfaces makes the classification more annoying.  I haven't thought that through.