Here's something I tried, thanks to Adam's comment for the basic idea for $p \ge 3$. This answer is missing some details, comments suggesting improvements are welcome, as would be other answers that fill in the gaps. It also relies on the fact (?) that:
$$\forall 0 \neq a \in \mathbb{C}, 0 \neq b \in \mathbb{C}, 1 < m \in \mathbb{N} . \exists x . |a e^{i x} + b e^{i m x}| \neq 1$$
Let $F(z, c) = z^2 + c$ with $F^{p+1}(z, c) = F^p(F(z, c), c)$. Now the boundary of a hyperbolic component can be parameterized by $\theta \in \mathbb{R}$ by the solution of the equation system:
$$
F^p(z,c) = z \\
\frac{\partial}{\partial z}F^p(z,c) = e^{i \theta}
$$
Now the question reduces to showing $c$ is of the form $c = c_0 + r_0 e^{i \phi}$ where $c_0 \in \mathbb{C}$ and $r_0 \in \mathbb{R}$ are constants and $\phi \in \mathbb{R}$.
$F^p(z,c) = z$ defines a polynomial of even degree $P(z) = 0$, whose constant coefficient is the product of its roots and is a polynomial in $c$ of degree $2^{p-1}$. The roots include those of $F^q(z, c) = z$ where $q | p$. Also, $\frac{\partial}{\partial z}F^p(z, c) = 2^p \Pi z_k$ where the $z_k$ are the $p$ roots in the periodic orbit of the desired solution $z$ (all $z_k$ are roots of $F^p(z, c) = z$, the remaining roots have lower period).
Case $p = 1$:
$$ z^2 + c_0 + r_0 e^{i \phi} = z \\
\therefore z = \frac{1 \pm \sqrt{1 - 4(c_0 + r_0 e^{i \phi})}}{2} \\
\frac{\partial}{\partial z} = 2 z = e^{i \theta}
$$
Now $|e^{i \theta}| = 1$ but $\exists x . |2 z| = |1 \pm \sqrt{x}| \neq 1$, so conclude that period $1$ component is not a perfect circle.
Case $p = 2$:
The equations reduce to
$$ 4(1 + c_0 + r_0 e^{i \phi}) = e^{i \theta} $$
with obvious solution $c_0 = -1, r_0 = \frac{1}{4}, \phi = \theta$, so conclude that the period 2 component is a perfect circle.
Case $p = 3$:
The equations reduce to
$$ 8 (c^3 + 2c^2 + c + 1) = e^{i \theta} \text{ where } c = c_0 + r_0 e^{i \phi}$$
For this to hold, the coefficients of $e^{i k \phi}$ must be zero for all $k > 1$. But setting $k = 3$ implies $r_0^3 = 0$ but we know that $r_0 > 0$ as hyperbolic components have non-empty interior. Contradiction, conclude that no period 3 component is a perfect circle.
Case $p > 3$:
Similarly to the $p = 3$ case, get a polynomial of degree $m > 1$ in $e^{i \phi}$ whose highest term has coefficient $r_0^m$. It remains to show that the polynomial really does have degree greater than $1$. The constant coefficient (product of roots) is a polynomial of degree $2^{p-1}$ in $c$, divided by the corresponding constant coefficient of all smaller divisors of the period gives:
$$m = 2^{p-1} - \sum_{q | p, q < p} 2^{q-1}$$
which solved numerically gives:
$$\begin{aligned}
p & & & m \\
1 & & & 1 \\
2 & & & 1 \\
3 & & & 3 \\
4 & & & 5 \\
5 & & & 15 \\
6 & & & 25 \\
\vdots
\end{aligned}$$
Finally, $m > 1$ for all $p \ge 3$ because $\exists q > 1 . q \nmid p$.
I am very new to algebraic number theory, so this may be complete garbage...
The first few iterations of $f_c^n(0)$, with their degree, are:
$$\begin{aligned}
0 && -\infty\\ c && 1 \\ c^2 + c &&2 \\ c^4 + 2c^3 + c^2 + c &&4 \\ c^8+4c^7+6c^6+6c^5+5c^4+2c^3+c^2+c&&8\\\vdots&&2^{n-1}\end{aligned}$$
Thus, Misiurewicz points $M_{q,p}$ of preperiod $q > 0$ and period $p > 0$ are roots of the monic integer-coefficient polynomial
$$f_c^{p+q}(0) - f_c^q(0)$$ and therefore each $c = M_{q,p}$ is an algebraic integer, with $\deg M_{q,p} \le 2^{p + q - 1}$. The algebraic integers are closed under addition and multiplication, therefore the multiplier
$$m =\prod_{n=q}^{p+q-1} 2 f_c^n(0)$$
is also an algebraic integer.
The desired multiplier is $(i \phi)^t, t\in \mathbb{R}$. $0 < t$ because Misiurewicz points are repelling, $|m|>1$. $i\phi$ is an algebraic integer of degree $4$, with minimal polynomial $x^4+3x^2+1$. Define $w(\cdot)$ to be the number of non-zero coefficients of the minimal polynomial of the multiplier, then $w(i\phi) = 3$.
Here is a table of some $w(M_{q,p})$:
q \ p 1 2 3 4 5 6
2 2 3 7 13 31 55
3 3 4 13 25 61
4 6 9 22 49
5 13 15 49
6 28 33
7 59
Conjecture 1: $\deg M_{q,1} = 2^{q-1}-1$
Conjecture 2: $w(M_{q,1}) = 2^{q-1}-(q-2)$
Conjecture 3: $w(M_{q,p}) = O(\deg M_{q,p})$
Claim: $w(M_{q,p}) = 3$ occurs only for $M_{3,1}$ and $M_{2,2}$, and their multipliers have degrees $3$ and $2$ respectively.
Now it is enough (is it?) to consider lowest-terms $\frac{a}{b} = t \in \mathbb{Q}$. Now suppose there is an $m$ such that $$m^{\frac{1}{a}} = (i \phi)^{\frac{1}{b}}$$ The right hand side has minimal polynomial $x^{4b}+3x^{2b}+1$ which also has $3$ non-zero coefficients, and similarly the left hand side has the same number of non-zero coeffients as $m$. Both are monic.
Therefore, the only possible choices for $m$ are the $m$ corresponding to each $M_{3,1}$ or $M_{2,2}$. The first implies:
$$x^{3a} - 4x^{2a} + 16 = 0 = x^{4b} + 3x^{2b} + 1 $$
So $3a=4b$, ie $a = 4n, b = 3n, n > 0$ but because $a$ and $b$ are coprime that means $n = 1$ and the equations boil down to:
$$x^{12} - 4x^{8} + 16 = 0 = x^{12} + 3x^{6} + 1$$
Rearranging gives:
$$4x^{8} + 3x^{6} - 15 = 0$$
But this has no solutions in the algebraic integers due to the leading $4$.
Similarly, for $M_{2,2}$ we get
$$x^{2a} - 8x^{1a} + 32 = 0 = x^{4b} + 3x^{2b} + 1$$
where $2a = 4b$ implies $a = 2$ and $b = 1$ so
$$x^4 - 8x^2 + 32 = 0 = x^4 + 3x^2 + 1$$
Rearranging gives $11 x^2 - 31 = 0$ but this has no solution in the algebraic integers due to the leading $11$.
Therefore there is no such $m$, and thus there are no Golden spirals in the Mandelbrot set.
Best Answer
Edit: previous answer was misleading, sorry.
Here's how you can do it: consider the function $$P_n(c)=\prod_{z=f_c^n(z)} \left( (f_c^n)'(z)-e^{2i\pi p/q} \right)$$ Then clearly a parameter $c$ satisfies your property iff $P_n(c)=0$.
Now it is not clear at first glance that $P_n$ is a polynomial (or even continuous), but it is. Let $A:=\overline{\mathbb{Q}}$ be the field of algebraic numbers. Observe that $f_c^n(z)-z$ can be seen as a polynomial in $z$ with coefficients in $A[c]$ (actually even in $\mathbb Z[c]$), and let $Q_1$ denote that polynomial. Then $P_n(c)$ is the resultant of $Q_1$ and of the polynomial $Q_2(z):=(f_c^n)'(z)-e^{2i\pi p/q}$, also with coefficients in $A[c]$.
Therefore, by properties of the resultant, $P_n(c) \in A[c]$, so the zeroes of $P_n$ are algebraic numbers.
In fact, this should also prove that parameters with an $n$-cycle of multiplier 1 are algebraic integers.