Here is a small improvement I can suggest. First - start iterating from 0. Second - exit loop when the square of root candidate exceeds the number.

function squareroot(number) {
    for (var i = 0; i * i <= number; i++) {
        if (i * i === number)
            return i;
   }
   return number; // don't know if you should have this line in case nothing found
}

This algo will work in O(√number) time comparing to initial O(n) which is indeed performance improvement that you asked.

Edit #1

Just even more efficient solution would be to binary search the answer as @Spektre suggested. It is known that x² is increasing function.

function squareroot(number) {
    var lo = 0, hi = number;
    while(lo <= hi) {
         var mid = Math.floor((lo + hi) / 2);
         if(mid * mid > number) hi = mid - 1;
         else lo = mid + 1;
    }
    return hi;
}

This algo has O(log(number)) running time complexity.