[vivekfan]

Is the strong base considered a reactant or is it just above the arrow (as in present in the reaction?)

[StarvinMarvin]

What difference does it make if it's above an arrow or not. Strong bases are used for a reason, so you can answer the question for yourself.

[vivekfan]

So is having something over the arrow equivalent to having as a reactant? Why is there difference in notation?

[vivekfan]

Where does the strong base anion go in the reaction?

Because in the book it shows the alkyl halide+ strong base = alkene + acid

Does it matter that the base is not there?

[vivekfan]

Actually I'm a bit confused. Is the hydrohalic acid HX formed, where X is the halide, or does the base deprotonate the alkyl halide and form HB, where B is the base? Does X just remain an anion?

[Arctic-Nation]

Anything written above (or below) reaction arrows usually is a catalyst or solvent, though it can also be a reactant that is not the substrate (for example, in an alkylation reaction, 'RX' can be written above the arrow).

In your case, the strong base is strictly speaking not a catalyst, as it is consumed by the reaction, thus it can be considered a reactant.

As for E2 eliminations, the base deprotonates the alkyl halide in the alpha position, which is immediately followed by elimination of the halide anion. So you get BH⁺ and X^-. E2 reactions, by definition, only occur in strong base, thus HX is not involved as such.

[orgopete]

I disagree with the statement of your textbook (if accurate). I am posting a mechanism which shows the bond making and breaking steps. I argue that you would be better served by understanding the process (the mechanism) than the terminology. That is, in order for the terminology to have meaning, it must refer to a general or specific use of that terminology. If you know specific examples of reactions, it is easier to learn the terminology associated with that reaction. Thus, in substitution and elimination reactions, the group delivering electrons to the halide is a nucleophile or base. The difference in the reactions is whether the electrons react with a proton or carbon, in either case a simple attraction of opposite charges (Coulombs Law). 

In either case, the halide is (formally) released as its anion. The electrons associated with a halogen are too tightly held to become protonated in an elimination reaction.

A common question about elimination/substitution is how to predict which reaction will take place. That question is too complex to give an answer to here.