The answer is yes, always, if $\operatorname{char}(F)=0$ and never if $\operatorname{char}(F)\neq 0$.

Suppose $\Gamma \subset G=G(F)$ is a lattice and $F$ is a non-archimedean local field with $char (F)=0$. By Margulis's arithmeticity theorem, $\Gamma $ is commensurate to $G_0(O_S)$ where $G_0$ is an absolutely simple algebraic group defined over a number field $K$, $S$ is a finite set of places of $K$ including all the Archimedean ones, and for some place $v$ of $K$, we have $F\simeq K_v$ with an isomorphism $G(F)\simeq G_o(K_v)$. Since the projection of $\Gamma $ to $G_0(K_v)$ is to be discrete, we have $G_0 (K\otimes _{\mathbb Q} {\mathbb R})$ is compact. But, the map of $\Gamma $ in $G_0(K \otimes _{\mathbb Q} {\mathbb R})$ is faithful, and the latter compact real algebraic group has a faithful unitary finite dimensional representation (this is essentially the argument of @YCor in the comments).

Suppose $\operatorname{char}(F)>0$. Since the rank of $G$ over $F$ is at least two, by super-rigidity in finite characteristic (due to myself and also proved in Margulis' book later), every finite dimensional representation of $\Gamma $ into $\mathrm{GL}_n(\mathbb C)$ has finite image. Hence $\Gamma $ cannot have infinite image finite dimensional (unitary) representations.

The answer is yes, always, if $\operatorname{char}(F)=0$ and never if $\operatorname{char}(F)\neq 0$. It is easy to see that it is enough to prove this when $G$ is simply connected as an algebraic group; we then assume $G$ is simply connected.

Suppose $\Gamma \subset G=G(F)$ is a lattice and $F$ is a non-archimedean local field with $char (F)=0$. By Margulis's arithmeticity theorem, $\Gamma $ is commensurate to $G_0(O_S)$ where $G_0$ is an absolutely simple algebraic group defined over a number field $K$, $S$ is a finite set of places of $K$ including all the Archimedean ones, and for some place $v$ of $K$, we have $F\simeq K_v$ with an isomorphism $G(F)\simeq G_o(K_v)$. Since the projection of $\Gamma $ to $G_0(K_v)$ is to be discrete, we have $G_0 (K\otimes _{\mathbb Q} {\mathbb R})$ is compact. But, the map of $\Gamma $ in $G_0(K \otimes _{\mathbb Q} {\mathbb R})$ is faithful, and the latter compact real algebraic group has a faithful unitary finite dimensional representation (this is essentially the argument of @YCor in the comments).

Suppose $\operatorname{char}(F)>0$. Since the rank of $G$ over $F$ is at least two, by super-rigidity in finite characteristic (due to myself and also proved in Margulis' book later), every finite dimensional representation of $\Gamma $ into $\mathrm{GL}_n(\mathbb C)$ has finite image. Hence $\Gamma $ cannot have infinite image finite dimensional (unitary) representations.

The answer is yes, always, if $char (F)=0$ and never if $char (F)\neq 0$.

Suppose $\Gamma \subset G=G(F)$ is a lattice and $F$ is a non-archimedean local field with $char (F)=0$. By Margulis's arithmeticity theorem, $\Gamma $ is commensurate to $G_0(O_S)$ where $G_0$ is an absolutely simple algebraic group defined over a number field $K$, $S$ is a finite set of places of $K$ including all the Archimedean ones, and for some place $v$ of $K$, we have $F\simeq K_v$ with an isomorphism $G(F)\simeq G_o(K_v)$. Since the projection of $\Gamma $ to $G_0(K_v)$ is to be discrete, we have $G_0 (K\otimes _{\mathbb Q} {\mathbb R})$ is compact. But, the map of $\Gamma $ in $G_0(K \otimes _{\mathbb Q} {\mathbb R})$ is faithful, and the latter compact real algebraic group has a faithful unitary finite dimensional representation (this is essentially the argument of @YCor in the comments).

Suppose $char (F)>0$. Since the rank of $G$ over $F$ is at least two, by super-rigidity in char p (due to myself and also proved in Margulis' book later), every finite dimensional representation of $\Gamma $ into $GL_n(\mathbb C)$ has finite image. Hence $\Gamma $ cannot have infinite image finite dimensional (unitary) representations.

The answer is yes, always, if $\operatorname{char}(F)=0$ and never if $\operatorname{char}(F)\neq 0$.

Suppose $\Gamma \subset G=G(F)$ is a lattice and $F$ is a non-archimedean local field with $char (F)=0$. By Margulis's arithmeticity theorem, $\Gamma $ is commensurate to $G_0(O_S)$ where $G_0$ is an absolutely simple algebraic group defined over a number field $K$, $S$ is a finite set of places of $K$ including all the Archimedean ones, and for some place $v$ of $K$, we have $F\simeq K_v$ with an isomorphism $G(F)\simeq G_o(K_v)$. Since the projection of $\Gamma $ to $G_0(K_v)$ is to be discrete, we have $G_0 (K\otimes _{\mathbb Q} {\mathbb R})$ is compact. But, the map of $\Gamma $ in $G_0(K \otimes _{\mathbb Q} {\mathbb R})$ is faithful, and the latter compact real algebraic group has a faithful unitary finite dimensional representation (this is essentially the argument of @YCor in the comments).

Suppose $\operatorname{char}(F)>0$. Since the rank of $G$ over $F$ is at least two, by super-rigidity in finite characteristic (due to myself and also proved in Margulis' book later), every finite dimensional representation of $\Gamma $ into $\mathrm{GL}_n(\mathbb C)$ has finite image. Hence $\Gamma $ cannot have infinite image finite dimensional (unitary) representations.