Since 1 and 4 are both residues (for any $p\ge 5$), then to avoid having consecutive residues (with 1 and 4), we would have to have both 2 and 3 as non-residues, and then we have 2 consecutive non-residues.
Thus, we must have either 2 consecutive residues or 2 consecutive nonresidues.
i.e.: 1 and 4 are both residues, so we have R * * R for the quadratic character of 1, 2, 3 and 4. However we fill in the two blanks with Rs or Ns, we will get either 2 consecutive Rs or 2 consecutive Ns.
Edited:
To show that we must actually get both RR and NN for $p\gt 5$, we consider 2 cases:
$p\equiv -1 \pmod 4$: then the second half of the list of $p-1$ Ns and Rs is the inverse of the first half (Ns become Rs and the order is reversed), so that if we have NN or RR in the first half (using the argument above) then we get the other pattern in the second half.
$p\equiv 1 \pmod 4$: then the second half of the list is the reverse of the first half. Then if there is no RR amongst the first 4, then there must be an appearance of NN, i.e. sequence begins RNNR..., and if we fill in the dots (this is where we need $p>5$ - to ensure there ARE some dots!) with Ns and Rs trying to avoid an appearance of RR, then we have to alternate ...NRNR...NR. However the sequence then ends with R, and the second half begins with R, so we eventually get RR.
(The comments about the second half of the list in the 2 cases are easy consequences of -1 being a residue or a nonresidue of p).
When $p=3 \pmod 4$, there is still a simple formula for adding all of the primitive roots.
There probably is no easy answer here. The primitive roots of 7 are 3 and 5, adds to 8, and for 11, one has 2, 6, 7 and 8, which adds to 23.
One can establish that the primitive roots add up to a number modulo $p$, specifically $(-1)^m$, where $m$ is the number of distinct prime divisors of $p-1$, and if $p-1$ is divisible by a power of a prime, then the sum of primitive roots is a multiple of $p$.
The proof of this is fairly straight forward. Consider for example, $30$. The sum of the complete solutions of $x^n \pmod p$ is $0$, where $n \mid p-1$
So one works through the divisors. For $n=1$, the sum is 1. For $n=p_1$, q prime the sum is -1. For $n=p_1 p_2$, the sum is +1. The sum of all of the divisors of $p_1 p_2$, is then $f(1)+f(p_1)+f(p_2)+f(p_1 p_2) = 0$.
For powers of $p_n$, the sum $f(1)+f(p_1)+f(p^2_1) \dots$, is zero, so every term after the second must be zero.
Since the primitive roots is the largest divisor of $p-1$, then it is by that formula.
Best Answer
Yes, there's a formula, though not quite as simple: the sum is $$ \frac12 \left( {q \choose 2} - \frac2w qh \right), $$ where $h$ is the class number of the quadratic imaginary field ${\bf Q}(\sqrt{-p})$, and $w$ is the number of roots of unity in that field, so that the factor $2/w$ is just $1$ except for $q=3$ when $2/w = 1/3$.
This is obtained by writing the sum of the residues as $\frac12 \sum_{r=1}^{p-1} (1+(r/p)) r$, where $(r/p)$ is the Legendre symbol. The $\sum_r r$ part of this gives $q \choose 2$, and the formula $-2qh/w$ for the $\sum_r (r/p) r$ part is classical (the factor $2/w$ arises via the Dirichlet class number formula, see for instance the $d<0$ part of equation (2) in MathWorld's "Class Number" page).
[The formula for $q \equiv 1 \bmod 4$ is simple because in that case the identity $(r/p) = (-r/p) = ((p-r)/p)$ yields $\sum_{r=1}^{p-1} (r/p) r = 0$ by symmetry.]