I will start with few observations; along the way, I will correct your question. Let $g\in SO(n,1)$ be an orientation-preserving loxodromic element. Let $U_g\in SO(n)$ denote the rotational part of $g$. Then $U_g$ always has a nonzero fixed vector in ${\mathbb R}^n$. Furthermore, if $n$ is even then it follows that the dimension of the fixed-point set of $U_g$ is $\ge 2$. Define $\delta_n=1$ if $n$ is odd and $\delta_n=2$ if $n$ is even. Let's call $g$ "strictly loxodromic" if dimension of the fixed point set of $U_g$ is exactly $\delta_n$. Clearly, $SO(n,1)$ contains orientation-preserving strictly loxodromic elements.
Let $\rho=\rho_g: {\mathbb Z}\to SO(n,1)$ denote the representation sending the generator to loxodromic $g$. Then dimension of the fixed-point set of $U_g$ is nothing but $b^0_\rho= dim H^0({\mathbb Z}; R^{n,1}_\rho)$. Now, it is a general fact that $$ \{\rho: {\mathbb Z}\to SO(n,1): b^0_\rho\ge m\} $$ is an algebraic subvariety of $Hom({\mathbb Z}, SO(n,1))=SO(n,1)$. The same holds if you consider $$ dim H^i(\Gamma, V_\rho) $$ where $G$ is an algebraic group, $\Gamma$ is a finitely-generated group and $V$ is a finite-dimensional linear algebraic representation of $G$ (just $Hom(\Gamma, G)$ will be more complicated in this case). In our setting, this is especially easy to see since $$ b^0_{\rho_g}= dim Ker (g-I), $$ where I regard $g$ as a matrix in $SO(n,1)$.
In particular, $$ \{g\in SO(n,1): b^0_{\rho_g}\ge \delta_n+1\} $$ is a (proper) algebraic subvariety. In particular, if $\Lambda < SO(n,1)$ is a Zariski dense torsion-free subgroup, then it cannot consist entirely of elements with $b^0_{\rho_g}\ge \delta_n+1$. In other words, such $\Lambda$ always contains a strictly loxodromic element, thus, answering the corrected version of your question.