It looks like your rows are sorted by timestamps (I will assume this for the rest of this post), so if there are duplicates, they will be next to each other (could be more than 2 rows). All what you need to do is (the algorithm):

open the file;
open a temp file for output;
set $lastrow to '';
while (file not empty) {
    read one row;
    if (this row equals to the $lastrow) {
         do nothing;
    } else {
         write this row to the output file;
         set $lastrow to this row;
    }
}
close both files;
copy the temp file to the original file;
[download]

The perl code would be close to this:

use strict;
use warnings;

my $lastrow = "";
while (my $line = <DATA>) {
    $line =~ /(.*?)\n/;
    $line = $1;
    if ($line ne $lastrow) {
        print $line, "\n";
        $lastrow = $line;
    }
}
__DATA__
data_row_051126120432.data 
data_row_051126120630.data 
data_row_051126120630.data 
data_row_051126122305.data 
data_row_051126122305.data 
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This prints:

data_row_051126120432.data
data_row_051126120630.data
data_row_051126122305.data
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It is not whether you "depend on the files being sorted", but whether it is a fact that the file is sorted.

If it is (for example it could be some sort of log), to hold the entire file in momery is then a waste.

my %seen = ();
while(<>) {
     print if ! $seen{$_};
     $seen{$_} = 1;
}
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#!/usr/bin/perl
use warnings;
use strict;
use Tie::File;

tie @file,'Tie::File','myfile' or die "Can't tie file: $!";
undef %saw;
@file = grep(!$saw{$_}++, @file);
untie @file;
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perl -pi -e'$_ = "" if $seen{ $_ }++' temp/*
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Makeshifts last the longest.