I am rewriting my comment as an answer. That is false in characteristic $p$ for torsors for the finite, flat group scheme $\mu_p=\text{Spec}\ \mathbb{Z}[t]/\langle t^p -1 \rangle$ with the usual group multiplication, $$\mathbb{Z}[t]/\langle t^p-1 \rangle \to \mathbb{Z}[t_1,t_2]/\langle t_1^p-1,t_2^p-1\rangle, \ \ t\mapsto t_1t_2.$$ Let $k$ be an algebraically closed field of characteristic $p$. Let $R$ be $k[[x]]$. Let $T$ be the $R$-scheme $$T=\text{Spec} \ S, \ \ S=R[y]/\langle y^p-(1+x) \rangle.$$ Let $m:\mu_p\times T \to T$ be the $R$-morphism given by $$m^*:S\to S[t]/\langle t^p-1 \rangle, \ \ y \mapsto yt.$$ I claim that this is a $\mu_p$-torsor for the fppf topology. Indeed, the $R$-scheme $T$ is itself fppf. We need to check that the induced map, $$(m,\text{pr}_2):\mu_p \times T \to T\times_{\text{Spec}\ R} T,$$ is an isomorphism. On the level of rings, this is given by the $S$-algebra homomorphism, $$S[y_1]/\langle y_1^p-(1+x) \rangle \to S[z]/\langle z^p -1 \rangle, \ \ y_1 \mapsto yz.$$ The inverse isomorphism sends $z$ to $y_1 y^{p-1}(1+x)^{-1}$. Thus, $T$ is a $\mu_p$-torsor over $\text{Spec} \ R$.

The reduction of this torsor modulo $x$ is just $\text{Spec}\ k[y]/\langle y^p -1 \rangle$, which equals $\mu_{p,k}$. However, there is no trivialization over all of $\text{Spec}\ R$. Indeed, for every $f\in R$, the element $(1+xf)^p$ equals $1+x^pf^p$, which cannot equal $1+x$. Thus, the torsor $T$ is nontrivial over $\text{Spec}\ R$.

When $G$ is smooth over $\text{Spec}\ R$, for torsors $T$ and $T'$ over $\text{Spec}\ R$, also the "inner twist" of $G$ by $T$ is smooth over $\text{Spec}\ R$, and the Isom scheme $\text{Isom}_R(T,T')$ is a torsor for this smooth group $R$-scheme. For a torsor for a smooth group scheme over $\text{Spec}\ R$, every $k$-point lifts to an $R$-point by Hensel's Lemma.