If $p(z) = a \prod_{j=1}^d (z - r_j)$ and $q(z) = b \prod_{j=1}^e (z - s_j)$,
$r_1 \le r_2 \le \ldots \le r_d$, $s_1 \le s_2 \le \ldots \le s_e$,
then let $$ f(z) = \dfrac{p'}{p} - \dfrac{q'}{q} = \sum_{j=1}^d \dfrac{1}{z-r_j} - \sum_{j=1}^e \dfrac{1}{z-s_j} = \sum_{k=0}^\infty \dfrac{m_k}{z^{k+1}}$$
for $z > R = \max( r_d, s_e)$, where $m_k = \sum_{j=1}^d r_j^k - \sum_{j=1}^e s_j^k$. We may assume wlog the sets $\{r_j\}$ and $\{s_j\}$ are disjoint.
The sign of $f(z)$ as $z \to R+$ is positive if $r_d > s_e$, negative if $r_d < s_e$. The sign as $z \to +\infty$ is that of the first nonzero coefficient $m_k$ (which must exist unless $d=e=0$). A sufficient condition for the existence of a zero of $f$ in $(R,\infty)$ is that
these signs are different.
Moreover, by bounding the tail of the series we can
place bounds on the location of any zeros. Thus suppose the first nonzero coefficient is $m_k$. We have
$|m_j| \le (d+e) M^j$ where
$M = \max(|r_1|,\ldots,|r_d|,|s_1|,\ldots,|s_e|)$, so that for $z > M$,
$$|f(z)| \ge \dfrac{|m_k|}{z^{k+1}} - \sum_{j=k+1}^\infty (d+e) \dfrac{M^j}{|z|^{j+1}} = \dfrac{|m_k|}{z^{k+1}} - (d+e) \dfrac{M^{k+1}}{|z|^{k+2}(1 - M/|z|)} $$
so any zero must satisfy
$$ z \le M + \dfrac{(d+e) M^{k+1}}{|m_{k}|}$$
EDIT: $m_0 = d - e$, so that's the first nonzero coefficient if $d \ne e$.
If $d = e$, then at least one of $m_1, \ldots, m_d$ is nonzero. To see this, consider the power-sums $M_i(X_1, \ldots, X_d) = \sum_{j=1}^d X_j^i$ for $ i=1\ldots d$ as symmetric polynomials in $X_1, \ldots, X_d$. It is well-known that these form an algebraic basis for the symmetric polynomials in $X_1, \ldots, X_d$. Let $Q(X_1,\ldots, X_d,y) = \prod_{j=1}^d (X_j - y)$. This is nonzero if and only if $y \notin \{X_1, \ldots, X_d\}$, and its $k$'th derivative with respect to $y$ is nonzero if and only if at most $k$ of $X_1, \ldots, X_d$ are equal to $y$. $Q(\cdot, \ldots, \cdot, y)$ and its derivatives with respect to $y$ are symmetric polynomials in $X_1, \ldots, X_d$, so they are determined by the power-sums $M_i(X_1, \ldots, X_d)$.
Thus the only way to have $m_j = M_j(r_1, \ldots, r_d) - M_j(s_1, \ldots, s_d) = 0$ for $j=1\ldots d$ is that $(r_1, \ldots, r_d)$ is a permutation of $(s_1, \ldots, s_d)$.
