I'm using four batteries with a total of 6 volts, but the lightstrip only needs five. How would resistance work?
LED strips designed for 5V will typically work with a power supply voltage from 3.5 to 5.5V. So there is a fair chance that with 6V (with new batteries it could even be as high as 6.6V), you could destroy the LED strip. A voltage divider as the one you describe, will not work, since the current through the resistors would have to be much too high. Even a solution with a simple series resistor would not work because the operating current is not constant. Next idea could be to use a Zener diode with a series resistor, but again, the current through the diode would cause a lot of power dissipation. 

A series diode (e.g. 1N5404) would reduce the voltage from 6V to about 5.3V (or at 2A rather 5V). However, battery voltage will decrease rapidly with time and at the end of the battery life time, the battery voltage will only be approximately 4V and the series diode would reduce that to about 3.3V. So you would loose a little bit of your battery capacity.

A low dropout voltage regulator could be another choice. However, it must be rated for the maximum operating current of the LED strip and the dropout should be well below 1V. Furthermore, a linear regulator will have a maximum dissipation power of about 2W.

Although it is technically possible to get 2A from a AA battery, the initial internal resistance is between 0.15 and 0.3 ohms per cell, which would lead to a voltage drop of 1.2 to 2.4V. In addition, the battery capacity at these currents is significantly less than 1Ah and, consequently, the battery would be empty very quickly. Therefore, it does not seem to be a sensible choice. I would recommend to use C (or D) batteries.