How to quote a variable in bash script

If should work as you have written it here (assuming that I'm parsing the garbled formatting correctly):

VARIABLE=`cat "file.txt" | grep "text"`
mycommand -p "$VARIABLE"

If there are problems handling special characters in the search results (other than null, which cannot be stored in a shell variable), then they're almost certainly due to mishandling in mycommand, not in how it's being called. Here's an example:

$ cat file.txt 
There are some $special \characters\\\\ in this text file.
This is another line of text.
$text $VARIABLE $HOME etc.
$ VARIABLE=`cat "file.txt" | grep "text"`
$ printargs -p "$VARIABLE"
Got 2 arguments:
    '-p'
    'There are some $special \\characters\\\\\\\\ in this text file.\nThis is another line of text.\n$text $VARIABLE $HOME etc.'

printargs is a simple python program that prints a pythonic representation of its arguments (note that it doubled all of the backslashes, and printed the line breaks as '\n', because that's how you'd write them in a python string):

#!/usr/bin/python

import sys

print "Got", len(sys.argv)-1, "arguments:"
for arg in sys.argv[1:]:
    print "    " + repr(arg)

BTW, I have a couple of stylistic suggestions: use lowercase names for shell variables to avoid conflicts with environment variables that have special meanings (e.g. assigning to $PATH will lead to trouble). Also, cat file | grep pattern is a useless use of cat; just use grep pattern file.