Case 1: none of p,q,r are congruent to 0 mod 3.  Then the equation mod 3 reduces to +/-1 = 0. This is a contradiction, so no solutions here.

Case 2: all of p,q,r are congruent to 0 mod 3.  Then needing all variables to be prime means the equation can only be 3^3 = 3^2 + 3^2 + 3^2.  This checks out.  

Case 3: two of p,q,r are congruent to 0 mod 3

Subcase 1: p=0 mod 3.  Then without loss of generality let q=3; then 27 = 9 + 9 + r^2 but then r=3 which is the Case 2 solution

Subcase 2: q and r are 0 mod 3.  Then p^3 = p^2 + 9 + 9.  Then we have p^3 - p^2 - 18 = 0, but the one integer root is p=3, which again is already covered in Case 2.

Case 4: exactly one of p,q,r are congruent to 0 mod 3

Subcase 1: p=0 mod 3.  Then 27 = 9 + q^2 + r^2, which then means 18 = q^2 + r^2.  This is very easy to check that the only positive integer solution is again q=r=3

Subcase 2: without loss of generality q=3.  Then p^3 = p^2 + r^2 + 9.  Now we'll go mod 6.  Then the equation reduces to +/1 = 4 +/- 1.  This is a contradiction, so no solutions in this case.

That leaves us with exactly one triplet (p,q,r) of prime numbers that satisfy the given equation is (3,3,3).