Select row at the nth percentile
Use the window function ntile() in a subquery (requires Postgres 8.4 or later).
Then select the segments you are interested in (corresponding to percentiles) and pick the row with the lowest value from it:

SELECT DISTINCT ON (segment)
       the_date, to_char((segment - 1)/ 10.0, '99.9') AS percentile, ans
FROM  (
    SELECT t1.the_date 
          ,ntile(1000) OVER (ORDER BY (t2.latency - t1.latency)) AS segment
          ,(t2.latency - t1.latency) AS ans
    FROM   table1 t1
    JOIN   table2 t2 ON t1.id = t2.id
   ) sub
WHERE  segment IN (601, 901, 991, 1000)
ORDER  BY segment, ans;


The Postgres-specific DISTINCT ON comes in handy for the last step. Detailed explanation in this related answer on SO:
Select first row in each GROUP BY group?

To get the 90, 99 and 99.9 percentile I picked the matching granularity with ntile(1000). And added a 60 percentile as per comment.

This algorithm picks the row at or above the exact value. You can add a line to the subquery with percent_rank() to get the exact relative rank of the select row in addition:

 percent_rank() OVER (ORDER BY (t2.latency - t1.latency)) AS pct_rank


Aside: I replaced the column name date with the_date since I am in the habbit of avoiding reserved SQL key words as identifiers, even if Postgres would permit them.