First, we show that if all roots of $Q$ have absolute value at most $1$, then $Q$ has a primordial multiple. We use induction on $\deg Q$.

If $Q$ is linear, then it is clearly primordial, so we are done.

Now assume $\deg Q > 1$, and let $a$ be a root of $Q$, and write $Q(x) = (x - a)Q_1(x)$. By induction hypothesis, $Q_1$ has some primordial multiple $P_1$, say of degree $d$. Then, $x - a \mid x^{d+1} - a^{d+1}$, so the polynomial $(x^{d+1} - a^{d+1})P_1(x) = x^{d+1}P_1(x) - a^{d+1}P_1(x)$ is a multiple of $Q$, and it is primordial: Indeed, coefficients of both $x^{d+1}P_1(x)$ and $a^{d+1}P_1(x)$ have absolute values at most $1$ (since $|a| < 1$), and the polynomials have no terms in common, so their difference is also primordial. So we are done by induction.

Now we show that $r > 1$ doesn't work. Choose a positive integer $d$ such that $r^d > 2$, and consider $Q(x) = x^d - r^d$. All roots of $Q$ have absolute value exactly $r$. Suppose $Q$ has a primordial multiple $P(x) = x^n + a_{n-1}x^{n-1} + \dots + a_0$, and let $a_n = 1$ for convenience. Let $\omega$ be the primitive $d$th root of unity. Consider the quantity

$$\frac{\sum_{i=0}^{d-1} \omega^{-ni} P(\omega^i x)}{d} = \sum_{j=0}^{n} a_j x^j \cdot \frac{(1 + \omega^{j-n} + \omega^{2(j-n)} + \dots + \omega^{(d-1)(j-n)})}{d} = \sum_{\substack{0 \le j \le n \\ j \equiv n \pmod d}} a_j x^j,$$

by standard roots of unity filter. Hence, if $l$ is the remainder that $n$ leaves upon division by $d$, the above polynomial is $x^l P_0(x^d)$ for some primordial polynomial $P_0(t) = t^m + b_{m-1}t^{m-1} + \dots + b_0$. But, since $\omega^{ir}$ are roots of $Q$ for $0 \le i \le d-1$, they are roots of $P$, so $P_0(r^d) = 0$ (because $r \ne 0$). But then,

$$\begin{aligned} r^{md} &= | - b_{m-1}r^{(m-1)d} - \dots - b_1r^d - b_0 | \\ &\le r^{(m-1)d} + \dots + r^d + 1 \\ &= \frac{r^{md} - 1}{r^d - 1} \\ &< r^{md} - 1 \end{aligned}$$

since $r^d > 2$, contradiction! Hence $Q$ does not have a primordial multiple, as required.