Powering 5v LEDs with 6 or 12 volt batteries
Power

According to the WS2812 datasheet (I can't find a WS2812B datasheet and you didn't provide a link) the integrated LED chip and controllers run on a supply voltage of 6 to 7 V and a current of 20 mA per R, G and B LED. Total all-on power demand given by \$ P = VI \$ will be \$ 2000 \cdot 3 \cdot 6 \cdot 0.02 = 720 W \$. You have underestimated on your power budget.

Current

Max current is simply the sum of all the currents when all LEDs are on. \$ I_{MAX} = 2000 \cdot 3 \cdot 0.02 = 120 A \$.

Battery

Batteries are typically rated in Ah (ampere-hours). The manufacturer will provide an amp-hour rating at a specific discharge rate. If you discharge at a higher rate the amp-hour rating decreases.

Ignoring that for a moment, at 6 V (no voltage conversion) you need 120 A x 16 h = 1920 Ah capacity. Let's round it up to 2000 Ah. We can de-rate, for example, if we assume that on average 50% of the LEDs will be on, to 1000 Ah.



As an example, the 6 V Yuasa battery pictured above has a 12 Ah capacity. You would need \$ \frac {1000~Ah}{12~Ah} = 83 \$ to give 1000 Ah capacity. Weight: 83 x 2.05 kg = 171 kg. Guide price: 83 x €16 = €1,328 from Farnell.

Charging

Charge current for an ideal battery will be \$ \frac {Ah~capacity}{charge~time} \$. For an 8-hour charge time that would work out at \$ \frac {1000}{8} = 125~A \$.

Comments

These are high currents at low voltages. A 100 to 200 A, 6 V lead acid battery charger will be difficult to find. 


You could split the power circuits into smaller independent battery circuits with individual chargers for each. 
All negatives could be connected together along with the data GND but the individual battery circuits should not share a common positive. 
Limit the number of LEDs in each chain to a number that the wiring can support.
Derate everything in my calculations by some further safety factor.