There are multiple ways:

• String.valueOf(number) (my preference)
• "" + number (I don't know how the compiler handles it, perhaps it is as efficient as the above)
• Integer.toString(number)

Things I like about your code

• The idea to calculate the length of a number with the logarithm is really good!
• In my opinion you are writing good comments.
• Good variable names
• Works as intended, without (to my knowledge) any bugs.

Criticism

returnDigitString()

• It is considered bad practice to put more than one command into one line. So please make line breaks after every ";".
• Your solution is pretty long (over 30 lines) in comparison to the complexity of the problem. You could also have done something like that:

    public static String returnDigitString(int digit) {
        String res = "";
        String[] digits = {"0", "1", "2", "3", "4", "5", "6", "7", "8", "9"};
        for(int i = 0; i <= 9; i++) {
            if(digit == i) {
                res += digits[i];
                break;
            }
        }
        return res;
    }

main()

• You are not using the array "reverseStr". The String "digits" is not used either.
• When I started your program the first time, I didn't know what to do, because your program didn't tell me. Before scanning a user input, I would tell the user to input something.

System.out.println("Please enter number:");
Scanner scn = new Scanner(System.in);
int number = scn.nextInt();

If you want to improve this point even further (which is highly recommended!), you can use something like that (you will have to use import java.util.InputMismatchException;):

System.out.println("Please enter number:");
Scanner scn = new Scanner(System.in);
int number;
while(true) {
    try {
        number = scn.nextInt();
        break;
    }
    catch(InputMismatchException e) {
        System.out.println("That's not a number!");
        scn.nextLine();
    }
}

This will check, whether the user really enters a number. If the user enters something else, the program will ask him again to enter a number.

• Something like that is considered bad practice:

if(number != 0) {
length = ( int ) (Math.log10(number) + 1 );}

Please write

if(number != 0) {
        length = (int) (Math.log10(number) + 1);
}

instead.

• "valStr" is not necessary. You can just write:

strSeq = returnDigitString(remainder) + strSeq;

But this really is a minor point and just my personal opinion. It's fine to use an extra variable for this.

Codestructure

• I would use an extra method for the content of the main-method. Just use the main-method to call the new method.

String myString = "1234";
int foo = Integer.parseInt(myString);

If you look at the Java documentation you'll notice the "catch" is that this function can throw a NumberFormatException, which you can handle:

int foo;
try {
   foo = Integer.parseInt(myString);
}
catch (NumberFormatException e) {
   foo = 0;
}

(This treatment defaults a malformed number to 0, but you can do something else if you like.)

Alternatively, you can use an Ints method from the Guava library, which in combination with Java 8's Optional, makes for a powerful and concise way to convert a string into an int:

import com.google.common.primitives.Ints;

int foo = Optional.ofNullable(myString)
 .map(Ints::tryParse)
 .orElse(0)