Part of the difficulty might be that even inside single quotes (or quoting construct as here) the backslash is a tricky customer.

knoppix@Microknoppix:~$ perl -E '
> $bp = q{\\dirA\dirB\dirC};
> say $bp;'
\dirA\dirB\dirC
knoppix@Microknoppix:~$ perl -E '
$bp = q{\\\dirA\dirB\dirC};
say $bp;'
\\dirA\dirB\dirC
knoppix@Microknoppix:~$ perl -E '
$bp = q{\\\\dirA\dirB\dirC};
say $bp;'
\\dirA\dirB\dirC
knoppix@Microknoppix:~$ perl -E '
$bp = q{\\\\dirA\\dirB\\dirC};
say $bp;'
\\dirA\dirB\dirC
knoppix@Microknoppix:~$
[download]

Riales has given you the best method but, just to show another way, you could split the paths rather than substituting the backslashes and shift off the common elements before joining what's left to get the relative path.

knoppix@Microknoppix:~$ perl -E '
> $bp = q{\\\\dirA\\dirB\\dirC};
> say $bp;
> $cp = q{\\\\dirA\\dirB\\dirC\\dirD\\dirE};
> say $cp;
> @bpParts = split m{\\}, $bp;
> @cpParts = split m{\\}, $cp;
> do { shift @bpParts; shift @cpParts } while @bpParts;
> $rp = join q{\\}, @cpParts;
> say $rp;'
\\dirA\dirB\dirC
\\dirA\dirB\dirC\dirD\dirE
dirD\dirE
knoppix@Microknoppix:~$
[download]

I hope this is helpful.