I took the solution from Jesper and added some aggregation to it on multiple run of the same code
def time[R](block: => R) = {
    def print_result(s: String, ns: Long) = {
      val formatter = java.text.NumberFormat.getIntegerInstance
      println("%-16s".format(s) + formatter.format(ns) + " ns")
    }
    var t0 = System.nanoTime()
    var result = block    // call-by-name
    var t1 = System.nanoTime()
    print_result("First Run", (t1 - t0))
    var lst = for (i <- 1 to 10) yield {
      t0 = System.nanoTime()
      result = block    // call-by-name
      t1 = System.nanoTime()
      print_result("Run #" + i, (t1 - t0))
      (t1 - t0).toLong
    }
    print_result("Max", lst.max)
    print_result("Min", lst.min)
    print_result("Avg", (lst.sum / lst.length))
}
Suppose you want to time two functions counter_new and counter_old, the following is the usage:
scala> time {counter_new(lst)}
First Run       2,963,261,456 ns
Run #1          1,486,928,576 ns
Run #2          1,321,499,030 ns
Run #3          1,461,277,950 ns
Run #4          1,299,298,316 ns
Run #5          1,459,163,587 ns
Run #6          1,318,305,378 ns
Run #7          1,473,063,405 ns
Run #8          1,482,330,042 ns
Run #9          1,318,320,459 ns
Run #10         1,453,722,468 ns
Max             1,486,928,576 ns
Min             1,299,298,316 ns
Avg             1,407,390,921 ns
scala> time {counter_old(lst)}
First Run       444,795,051 ns
Run #1          1,455,528,106 ns
Run #2          586,305,699 ns
Run #3          2,085,802,554 ns
Run #4          579,028,408 ns
Run #5          582,701,806 ns
Run #6          403,933,518 ns
Run #7          562,429,973 ns
Run #8          572,927,876 ns
Run #9          570,280,691 ns
Run #10         580,869,246 ns
Max             2,085,802,554 ns
Min             403,933,518 ns
Avg             797,980,787 ns
Hopefully this is helpful