No, we can't. Assume otherwise. Considering parity, we obtain that (at least) one of the primes involved must be $2$, and it obviously cannot be $t$ (LHS would be greater than RHS). Therefore, WLOG $p=2$. Now, consider cubic residues modulo $9$. Those are only $0$, $1$ and $-1$. This implies that (at least) one of the cubes in the equation must have a remainder of $0$ modulo $9$ (the residues on both sides can't otherwise match up). Note that for a number to have a remainder of $0$ modulo $9$, it must be divisible by $3$. Therefore, when it is a cube, the base must be divisible by $3$, but the base has to be a prime, so the base is $3$. Also, $t\neq3$ (otherwise LHS would be greater than RHS). Also, $p\neq3$ (because $p=2$). Therefore, WLOG $q=3$. Now, consider cubic residues modulo $7$. Those are also only $0$, $1$, and $-1$. Similarly to the previous paragraph, we obtain that (at least) one of the cubes has a remainder of $0$ modulo $7$ and therefore one of the primes is equal to $7$. Now we have 2 cases: either $t=7$, or $t\neq7$. Case $t=7$: We obtain $8+27+r^3+s^3=343$, which is equivalent to $r^3+s^3=308$. This implies that both $r$ and $s$ are lesser than $7$, but then $r^3+s^3\leq5^3+5^3=250$, therefore this case has no solutions. Case $t\neq7$: WLOG $r=7$. We obtain $8+27+343+s^3=t^3$, which is equivalent to $t^3-s^3=378$. Clearly $s<t$. With the exception of $2$ and $3$ ($s=2$, $t=3$ is not a solution) a difference of $2$ primes is at least 2. Therefore, $t^3-s^3\geq(s+2)^3-s^3=6s^2+12s+8$. This is obviously increasing for positive integers, and for $s=7$ it is already greater than $378$, so we only have to check $s=2$, $s=3$ and $s=5$, neither of which works. Therefore this case also has no solutions. Both of the cases have no solutions, so there are no solutions, as we wanted to prove.