Every quadratic residue is of the form $k^2$ for a $k \in \{1,\,\dotsc,\, \frac{p-1}{2}\}$. So the product of quadratic residues is
$$\prod_{k=1}^{\frac{p-1}{2}} k^2.$$
Now relate that to Wilson's theorem: Since $k^2 \equiv (-1)\cdot k \cdot (p-k) \pmod{p}$, we have
$$\prod_{k = 1}^{\frac{p-1}{2}}k^2 \equiv (-1)^{\frac{p-1}{2}}\prod_{k = 1}^{\frac{p-1}{2}} k \cdot \prod_{k = 1}^{\frac{p-1}{2}}(p-k) =(-1)^{\frac{p-1}{2}}\cdot (p-1)! \equiv (-1)^{\frac{p+1}{2}}\pmod{p}.$$
Thus the product of quadratic residues modulo $p$ is $\equiv 1 \pmod{p}$ if $p \equiv 3 \pmod{4}$ and it is $\equiv -1\pmod{p}$ if $p \equiv 1 \pmod{4}$.
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
The original question has been answered (they are always a cyclic group). Perhaps slightly more interesting is:
If there is a primitive root modulo $m$ (that is, if the group of units modulo $m$ is cyclic), then the property holds. This occurs when $m$ is a power of an odd prime, twice a power of an odd prime, $m=2$, or $m=4$.
It also holds if $m=8$, since then the only quadratic residue is $1$, so the group of quadratic residues is cyclic. In fact, it holds if $m=2^n$ is a power of $2$, the result holds: because the group of units modulo $2^n$ is isomorphic to $C_{2^{n-2}}\times C_2$ (where $C_k$ is the cyclic group of order $k$), so the group of squares is isomorphic to $C_{2^{n-3}}$, which is cyclic.
Other nontrivial examples include $m=3p^a$ or $m=6p^a$ where $p$ is an odd prime and $a\gt 0$, $m=35 = 5\times 7$ (or more generally, $m=5^a 7^b$); and many others.
To get the complete answer, factor $m$ into primes, $$ m = p_1^{a_1}\cdots p_r^{a_r}$$ where $p_1\lt p_2\lt\cdots\lt p_r$ are primes, and $a_r\gt 0$. By the Chinese Remainder Theorem, the group of units modulo $m$ is $$\left(\mathbb{Z}/m\mathbb{Z}\right)^* = \prod_{i=1}^r \left(\mathbb{Z}/p_i^{a_i}\mathbb{Z}\right)^*.$$
The group of units modulo $p_i^{a_i}$ is
The group of squares of invertible elements modulo $m$ is then isomorphic to a product of groups of the form
The product is cyclic if and only if the orders of the cyclic factors are pairwise relatively prime.
This gives:
Theorem. Let $m$ be a positive integer, and let $p_1,p_2,\ldots,p_r$ be the distinct prime divisors of $m$, $p_1\lt p_2\lt\cdots\lt p_r$. Let $\varphi$ be Euler's totient function. The subgroup of squares of the invertible elements modulo $m$ is cyclic if and only if:
It is straightforward now to show (e.g., using Dirichlet's theorem on primes in arithmetic sequences) that there are $m$ with arbitrarily many distinct prime divisors for which the subgroup of squares of the units modulo $m$ is cyclic.