Hi Shane and welcome to Electronics Point
The simple answer is that it depends on how much voltage the larger segments need, and what current (brightness) you want to run them at. From your description ("decent size scoreboard") I think the answer is probably no.
In that circuit, the LED displays are driven by CD4543 ICs. You can get the data sheet for this IC from various manufacturers: ON Semiconductor (
http://www.onsemi.com/pub_link/Collateral/MC14543B-D.PDF), Texas Instruments (
http://www.ti.com/lit/ds/symlink/cd4543b.pdf), NXP (
http://www.nxp.com/documents/data_sheet/HEF4543B.pdf), and STMicroelectronics (
http://www.st.com/web/en/resource/technical/document/datasheet/CD00000332.pdf). These ICs are powered from the main supply rail, which is nominally 12V but can be increased up to 15V without damaging the ICs.
The amount of current that the CD4543's outputs can drive into the display is shown by the following graph.
This graph is from the ON Semiconductor data sheet for the device and represents the typical, not guaranteed, characteristics.
Assuming you power the circuit from 15V (which will give the best performance), the best output drive happens at the intersection of the two curves that I've circled in red. This point indicates a current of about 17 mA and an output voltage (sinking) of about 4V, which leaves 11V available for the LED segment and its current limiting resistor.
So if each segment in your display will operate from 9V or less, and will be bright enough at an operating current of 17 mA, on average one of those ICs will be able to drive it directly.
If not, you will need some kind of buffering, using transistors or MOSFETs. With these, there is not really any limit on the amount of voltage and current you can switch.
Once you have decided what you want to use for your displays, post all the informtation here, including the forward voltage, the operating current, and the part numbers of the LEDs that you want to use, and I will draw up a suggestion for how to drive them.
