Counter counter

Python 3, 63

print(sum(map(input().count,"#0469@ADOPQRabdegopq$%&8B$%&8B")))

A straightforward approach. Iterates over each character with a closed counter, summing the number of occurrences, doing so twice for characters with two closed counters. It would be the same length to instead write

"#0469@ADOPQRabdegopq"+"$%&8B"*2

Python 3 is needed to avoid raw_input.

CJam, 41 39 37 34 bytes

"$%&8Badopq#0469@Rbeg"_A<eu+qfe=1b

Thanks to @jimmy23013 for golfing off 3 bytes!

Try it online.

How it works

"$%&8Badopq#0469@Rbeg"             e# Push that string.
                      _A<          e# Retrieve the first 10 characters.
                         eu+       e# Convert to uppercase and append.
                                   e# This pushes "$%&8Badopq#0469@Rbeg$%&8BADOPQ".
                            q      e# Read from STDIN.
                             fe=   e# Count the occurrences of each character. 
                                1b e# Base 1 conversion (sum).

Pyth, 31 bytes

sm@tjC"cúÁ-ÈN%³rØ|­"3Cdz

Demonstration.

Note that the code may not be displayed properly due to the use of non-ASCII characters. The correct code is at the link.

I made a lookup table of the output desired for each input character, rotated it by 32 to make use of Pyth's modular indexing, stuck a 1 at the beginning, and interpreted it as a base 3 number, giving the number 2229617581140564569750295263480330834137283757. I then converted this number to base 256 and converted it to a string, which is the string used in the answer.