Firstly the battery is not \$5\mathrm{V}\$, it is nominally \$3.7\mathrm{V}\$ as pointed out already.
However the confusion lies in misinterpreting the \$\mathrm{mAh}\$ rating of the battery, and it has nothing to do with converter losses, and everything to do with
$$P=I\times V$$
If the battery is rated at \$6\mathrm{Ah}\$, it means you can draw \$1\mathrm{A}\$ for \$6\mathrm{\space hours}\$. So lets say you are drawing \$1\mathrm{A}\$ from the battery. This means it is delivering:
$$P=I\times V = 1 \times 3.7 = 3.7\mathrm{W}$$
Now lets say you feed this through an ideal boost converter. In this case \$P_{in}=P_{out}\$. So assuming we boost it up to \$5\mathrm{V}\$, this means that the current that must be drawn from the output to discharge the battery at this rate is:
$$I=P/V=3.7/5=0.74\mathrm{A}$$
So what this means is at the higher voltage, you can draw a current \$0.74\mathrm{A}\$ for \$6\mathrm{\space h}\$ for that battery capacity - so the output capacity is \$4.44\mathrm{Ah}\$.
Again, this is not converter losses - in fact with losses the number would be lower. Instead it has to do with the fact that you are sacrificing current capacity for a voltage gain which is how a boost converter works - if this wasn't the case then you would have invented free energy.
Essentially at a higher voltage, the energy delivered by each unit of charge \$\left(\mathrm{V}=\mathrm{J}/\mathrm{C}\right)\$ is higher, hence at a higher voltage but lower current you are still delivering the same energy.
160mA at 6V is about 1W that would be easily suffucuent for a fairly bright LED path light.
If you're not equipped to measure the current consumption of the lights
maybe go for a 2A 6V supply that should be plenty if you estimate is in the right ballpark.
You could probably even use a 5V 2A "phone charger" as battery powered devices are usually designed to operate from as low as 1.1V per cell so 5V should be plenty.
Best Answer
It will probably be as easy as you are hoping.
Chances are, the red wire is positive and the black wire is negative.
Can't say for certain without knowing more about the LED string, but chances are if it could be powered with 3 AA batteries, a USB will have more than enough capacity. Hopefully there is an inline current limiting resistor to the LED's, and therefore the additional 0.5 V will be dissipated over that. A little more wasted power, but nothing to worry about.