Greg's & Peter's answers are trying to answer the question you've actually asked, but they're kind of missing the important thing, which is that you probably don't have enough voltage to do what you want to begin with.
You don't say what battery chemistry you're dealing with, so let's go through all the common cases:
These typically start at about 1.6V per cell when fresh out of the factory, drop a bit due to self-discharge to a nominal 1.5V by the time you buy them, then have significant power down to about 0.9 or 0.8 V per cell.
A "6V" alkaline battery will have 4 cells, because they're sold based on the typical shelf voltage when bought, not their peak voltage. This is also why alkaline batteries have started to come with "best if used by" dates on their label: this is the point where they are expected to have self-discharged too much.
The drop is fairly linear for this type of chemistry: the halfway point is roughly at (1.6 - 0.8) ÷ 2 = 0.4 V down from the peak voltage, or about 1.2 V.
Meanwhile on the load side, you have 6.0V - (5.0V - 5%) = 1.25V ÷ 4 cells ≈ 0.3V per cell dropped before the circuit starts to misbehave. (5% being a typical dropout value for 5V TTL circuits.) Since there is an 0.8V battery voltage range where the battery is delivering significant power, this means you've still got something like ⅝ of the power in the battery when your circuit dies.
You don't talk about details of your circuit, so maybe yours will actually run down to 0.8V per cell, or 3.2V total. But, realize that we haven't even started talking about the voltage drop from the series diode or resistor yet.
The story here is similar to alkaline, except that the discharge curve is flatter and the minimum cell voltage is highly dependent on conditions. (Discharge current, temperature, cell type, etc.)
Just for the sake of argument, let's say your battery runs from about 2.1V when fresh off the charger down to about 1.8V when dead under load. Since the nominal cell voltage for lead-acid is 2.0V, you have a 3-cell battery, so that our guess at the circuit's minimum operating voltage of 4.75 ÷ 3 ≈ 1.6V per cell.
That sounds great, right? We only need 1.6V per cell to keep our load circuit powered, but we're going to get all the power out of the battery by about 1.7V or 1.8V. But, we still have not considered the series diode or resistor. If we pre-drop the battery voltage by 0.7V with a diode, we shift the per-cell voltages down by about 0.2V each, so that now we're right on the borderline of getting all the power out of the battery before the circuit dies.
Bottom line, a lead-acid battery might be suitable here with the diode trick. You might not be able to afford the bulk and weight of such a battery, though.
If you're after small and light rechargeable batteries, you're going to skip right over lead-acid. Nickel chemistries have a much better energy density than lead acid, which is why they're used in portable power tools.
Both NiCd and NiMH behave similarly for our purposes here, so we can consider them together.
These have a nonlinear discharge curve: they drop off really fast when straight off the charger, level off quickly, then drop fairly slowly over most of the cell's useful life. Then near the end of the cell's life, they start dropping faster and faster until their terminal voltage falls off a cliff under load.
The useful voltage range for a nickel-chemistry cell is roughly 1.3 V when fresh off the charger down to 1.0 V per cell when nearly dead. The nominal cell voltage for nickel chemistries is 1.2V, so a "6V" nickel-chemistry battery will have 5 cells.
Since the cells are mostly dead at about 1.0V, the battery is mostly dead at 5.0V. There's a bit more juice there to squeeze out, but it's only going to be about 10% or less of the battery's useful power. 1.0V is on the "knee" of the discharge curve. There is much less battery life between 1.0V and 0.9V as between 1.1V and 1.0V.
What happens when we put a diode or resistor in the way to pre-drop this to something your circuit can handle? This effectively shifts the battery curve down by some amount. A diode would shift the useful battery voltage range down to 5.3 - 4.3V. Although the discharge curve is nonlinear, the midpoint of the battery's useful life is still in the middle of this new range, or about 4.8V, by which time the circuit is probably about to die due to voltage starvation.
Bottom line, this isn't really that much better in terms of battery use efficiency than alkaline.
There's a whole zoo of lithium-based chemistries.
If you actually have a lithium battery there, it's probably one of the 1.5V chemistries that more or less behaves the same as alkaline from a voltage profile standpoint, so you have your answer above.
I doubt you have one of the ~3-4V chemistries, as they tend to cluster around the middle of that range, so that a 2-cell lithium battery would probably be sold as a 7V battery rather than a 6V battery.
A Better Plan
Instead of trying to drop your 6V battery to 5V and hoping it stays there, I'd suggest a DC-DC converter, of the sort that can increase or decrease the battery voltage, as required. One such design is the buck-boost converter, but there are others, such as the SEPIC and split pi types.
The idea here is to get one that will run from as little as 3-4V input and which will tolerate at least 6-7V input, all while putting out a nominal 5.0V.
I did some searching on Digi-Key and found about a dozen suitable-looking alternatives. I don't want to list them here because the question is so vague that I can't be sure they're really suitable to your application. The point is, go do your own searching. This is so common a problem that you are sure to find something suitable. You don't need to reinvent the wheel here.