Can the BQ76930 DSG and CHG driver handle 8 FETs each?

Whether the bq76930 can suitably drive multiple FETs or any particular FET will depend on the FET(s) and the current switched.

The bq76930 is specified with a 10 nF load.  A FET selected will have a particular load.  A lighter gate load from the FET will allow faster switching, a heavier FET load will slow the switching.  Current in the FETs will also have an effect.  As you add FETs the switching will become slower from the Ciss of the added FETs.  FET drain current can cause some feedback where eventually the FETs will not switch suitably.

The IRFR7446 is a high current FET.  Ciss on the datasheet I found is 3.15 nF typical, so you may achieve close to the bq76930 datasheet swiching with 3 of the FETs.  For 8 FETs I expect you will need a driver.  Additionally note the CHG output turn off is weak, most of the FET turn off will be provided by the gate-source resistor.  This will typically work suitably for low FET counts and current but will likely need a driver for higher currents and loads.

When using an external buffer or FET driver, be sure the turn off is not so fast that it excites system inductances and causes high voltage transients. V = LxdI/dt, you want to control dt.  Check with your FET supplier for recommendations on individual gate resistors for the paralleled FETs.  For the charge FET remember that the driver must reference to PACK-, so watch for the supply voltage and a level shifter for the input.  Additionally recall that PACK- will rise to PACK+ or above in transient if there is a discharge fault, so be sure to accomodate that effect in the circuit.

Thanks for the info. Noted on the FET required soft-start and weak CHG output turn OFF.

Slower switching issue should not be a problem on steady state currents right?
The FETs should only be either ON or OFF.
Please correct me if I am wrong. :-)

Or do you mean to say that if the input capacitance is exceeded they (FETS) wont turn ON/OFF at all. :-(

The slower switching from the internal R of the driver and the sum of the Ciss of the FETs may not be a problem with no current in the FETs.  It may be come a problem with current flowing.  As the FETs switch slowly they have a time with a large RDS so the current is high and the voltage across the FET is also high causing high power dissipation in the FET.  Be sure to check the safe operating area curves and information in the FET datasheet. If the FET gets too hot it will fail. In the battery the highest current for switching is likely the short circuit current.  Switching off a load current due to over temperature, over current or under voltage may also be a high current.  Typically charge current is much lower in a system than discharge current so the slow turn off of the charge FETs is acceptable.  Systems which charge at high currents will need more attention to the charge driver.

At turn on when the driver is too weak, apparently there can be a point where the change in the drain voltage as the FET starts to turn on can feed through the Cgd portion of the Ciss to push the gate voltage down so that the FET turns off again before it is on.  With the driver still pulling on the gate, it tries to turn on again.  This effect repeats and the FET array may not come on. Some FET manufacturers recommend individual gate resistors to avoid the gate of one FET from affecting the others in the array. It would be good to check your FET vendor's literature or other educational resources to understand these effects better.