Using a step-up converter offers the advantage of being able to use partially-depleted batteries whose output voltage falls below the operating voltage of your device, but offers the disadvantages of exposing the device to harm if the battery voltage exceeds the expected level (e.g. if your 3.3-volt device operates off two AA batteries and have a 3-volt power-in jack into which someone plugs a 5-volt supply), and of increasing battery draw as the batteries get depleted (which in turn increases the risks that rechargeable batteries will be damaged, or non-rechargeable batteries will "leak" (i.e. ooze corrosive chemicals). It also, for better or for worse, will generally cause devices to continue to work normally as the batteries age, until they suddenly reach a point where they quit altogether.
Using a step-down converter offers greater immunity to input overvoltage, and will cause the voltage which is fed to the main circuitry to drop once the battery voltage has sagged too much. In many cases, this will cause the device to start working less well as the batteries age--sometimes a good thing, and sometimes a bad thing. Such behavior may be a bad thing if the device becomes useless as soon as its performance starts to degrade, but may be a good thing if the device remains somewhat useful, and if a user might not want to have to change batteries at unexpected times.
There are many solutions for 5V output. USB can typically take at least 5.25V, which means that 5.3V is pushing it perhaps too far for some very sensitive devices.
However, you don't need a 7805 for getting to 5V stable. There are many LDO regulators with 0.5A - 1.5A current ability that have much lower drop-out. A LF50AB for example can go from 5.3V to 5.0V up to 500 mA (which is the max draw on a standard USB 2.0 port.)
Another option is, as has been suggested, a switching converter of some sort. These are more efficient than linear regulators in many cases, although when going from 5.3V to 5.0V with a linear regulator, you're already a 94% efficiency, which is great in that case. A particular regulator I like because it's cheap and reasonably high current and efficiency is the Murata OKI 78SR. It takes 7-35 Volts in, and outputs 5.0V up to 1.5A, with the same pinout as a 7805, but uniformly high efficiency.
Finally, batteries are typically higher voltage than indicated when fresh, and lower voltage than indicated when depleted. Thus, using linear regulators is often not the right choice. Putting all the batteries in series and using a switching buck regulator (as suggested above) is a good way to get around this. Putting all the batteries parallel and using a boost converter is also often acceptable -- and for Lithium chemistries, easier to charge correctly. If your batteries start above the target voltage and end below the target voltage, you will need a buck/boost, SEPIC, or similar converter that can work with the input being both above, and below, the input voltage. Pololu has some ready-made ones, although they are not terribly high-current rated, and these topologies are often less efficient than buck-only or boost-only.
Best Answer
As it is a step down converter it will perform the voltage conversion with a certain efficiency which is dependent on the input voltage (among other things). A general rule of thumb is that the with an increased voltage difference your efficiency will decrease.
So for example if you step-down from 12V to 5V you will have an efficiency of 85%, if you step-down from 45V to 5V it will be 70% (those are not real numbers, just some values thrown in, the drop in efficiency will probably be much smaller for a decent design).
Leaving that effect aside for a moment, you will increase the runtime if you put another battery in series to your existing ones. Basically you increase the amount of energy available. The energy is transferred to the output with said efficiency. So you have 7.4V and (just guessing) 2000mAh at the input, that is 14.8Wh with 80% efficiency you will have roughly 12Wh at the output.
If you put another battery in series the voltage will increase to 11.1V, the capacity will stay at 2000mAh, but the energy is now 22.2Wh and with the reduced efficiency (75%) roughly 16.5Wh at the output. So you increased the energy available to your Raspberry and it will run longer.
That said to increase the runtime, it would be better to put another battery in parallel, it would increase the capacity but not the voltage, so the efficiency will stay at the value. So instead of 7.4V with 2000mAh you could use a second pack and get 7.4V with 4000mAh, doubling the energy and runtime.
A note on buck-boost topologies: the efficiency can be very different in the different modes of operation, to get best efficiencies you better stay out of boost region of operation and just in the buck mode. (So the input should always be larger than the output, a 7.4V pack is fine in that regard)
A note on battery packs: connecting lithium batteries in series requires balanced charging or they will fail prematurely (they start to drift apart and one cell will get damaged).
Connecting batteries in parallel should have every cell connected in parallel not just the packs (so each cell can balance itself with the partners).
It is recommended to use cells of the same type to build a pack.
A note on rechargeable batteries: do not overdischarge them, implement some kind of low battery cut-off.