Determining if two Strings have common subtrings of a given length in Java
Don't add unnecessary pre-checks
Your method getSubstringsSet starts with two early returns:
if(aString.length() &lt; lengthOfSubstring){
    return setOfSubstrings;
}

if(aString.length() == lengthOfSubstring){
    setOfSubstrings.add(aString);
    return setOfSubstrings;
}

You actually don't need them. The rest of the code handles both cases just fine and it doesn't add to clarity or improve performance. Early-returns should be used when the rest of the code can then assume a condition that was checked, making its code simpler; but it is not the case here.
Use built-in methods
To add the substrings, you are currently doing:
StringBuilder sb = new StringBuilder();

//add each substring of length (lengthOfSubstring) to the setOfSubstrings. 
for(int j = 0; j &lt; lengthOfSubstring; j++){
    sb.append(charArray[i + j]);
}

setOfSubstrings.add(sb.toString());

So you're creating a new StringBuilder and then appending each character. This is a complicated way of retrieving the substring between two indices. It also allocates a new StringBuilder each time which isn't really needed.
You can just have
setOfSubstrings.add(aString.substring(i, i + lengthOfSubstring));


In the same way, stringIntersect is a lot more complicated than it should be.
First of all, you are using a label outerForLoop and then using it to break out of the inner loop with break outerForLoop;. This is generally not a good practice. What this shows is a missing method: what you really want is to make a method isIntersection(set1, set2) that would determine if the two sets have elements in common.
However, you don't even need to write one, since it is already built-in: Collections#disjoint. This method returns true if the two given collections are disjoint, i.e. have no elements in common. Therefore, you can remove your double for loop and just have:
boolean sameSubstringInBothStrings = !Collections.disjoint(setOne, setTwo);