No, for a counterexample, just take $$ \nu(dx) = \left(C_1 x^{-2} I(0<x<1) +C_2 |x|^{-2} I(-1<x<0)\right)dx, $$ where $C_1\neq C_2$. Then, $$ \int_{r<|x|\le 1} x\nu(dx) = (C_2-C_1) \int_r^1 x^{-1} dx \to \infty,\quad r\to 0. $$ As Levy measure is $0$ for $|x|\le 1$, all moments of the Levy process $X_t$ are finite. The existence of the Levy process with this Levy measure is ensured by the references I gave earlier.
Your question seems to be connected to Levy processes of bounded variation. The following is Lemma 2.21 in the second edition of Kyprianou's book cited above: A Lévy process with Lévy–Khintchine exponent corresponding to the triple $(a,\sigma,\nu)$ has paths of bounded variation if and only if $$ \sigma=0 \mbox{ and } \int_R (1\wedge |x|) \nu(dx)<\infty. $$