Should "%%a" count as one (x=%) or two (x=%,x=a)? The solution below returns the former.

my $count = my @order = $string =~ /%(.)/sg;
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Or if you want the count for each different "x" instead of one global count:

$count{$_}++ for my @order = $string =~ /%(.)/sg;
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Update: Added second solution.

while ($x =~ m/%(.)/g) {
  push(@order, $1);
  $count{$1}++;
}
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use strict;
use warnings;
use Data::Dumper;

my @list = ("-----%s----%a----%d--%s-%d---" =~ m/\%./g);
print Dumper \@list;
__END__
$VAR1 = [
          '%s',
          '%a',
          '%d',
          '%s',
          '%d'
        ];
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my $s = "-----%s----%a----%d--%s-%d---";

print $s =~ tr/%//, "\n";   
 
print "$_\n" for ( $s =~ /%([ads])/g );

OUTPUT :
5
s
a
d
s
d
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To count them,

$_=q(aaa%bbbbb%cccccc%ddd);
print tr/%//, $/;
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To extract their positions;

$_ = 'aaa%bbbbb%cccccc%ddd';
print $-[0], ' ' while /%/g;
print $/'
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See perlvar for the @- array magic. You would push those values to an array, instead of printing, for further use.

The counting solution will give a wrong result if the string has a trailing %, because the OP wants the number of %s followed by another character.