Let be focus on the two-dimensional case first. Finding an integral model for $\rho$ is equivalent to choosing a lattice in $K^2$ that is invariant under $\rho(G)$. So in this case you are asking for invariant points in the tree for $GL_2(K)$ under a compact subgroup $\rho(G)$ of $GL_2(K)$. There are various possibilities, and there is a nice paper of Bellaiche and Chenevier Sous-groupes de $GL_2$ et arbres discussing the possibilites.

When you say that you would like all integral models to be semi-simple, I'm not sure what you mean; but it say the represenation over $K$ is irreducible, then it will never happen that all the integral models have semi-simple reduction. (This is a proposition of Ribet, and is easily seen in the tree-based picture.)

Just to illustrate how one argues: if two points in the tree are $\rho(G)$-invariant, then so will be all the points on the line segment joining them. If $\rho(G)$ fixes an infinite half-line, this corresponds to an invariant one-dimensional space over $K$, and so means that $\rho$ is reducible over $K$. So if $\rho(G)$ is irreducible over $K$, then the fixed set of $\rho(G)$ will be bounded and convex, which already imposes some restrictions. (One an prove Ribet's result by these sort of considerations: basically, any extremal point of the fixed set of $\rho(G)$ will have non-simple reduction.)

Once you understand the $GL_2$ case you can imagine how to try and generalize things to the $GL_n$ case, although the combinatorics becomes more complictated, and (as Bellaiche and Chenevier note) it is harder to say anything nice, because when $n > 2$, $GL_n$ is the full automorphism group of the building.

Let be focus on the two-dimensional case first. Finding an integral model for $\rho$ is equivalent to choosing a lattice in $K^2$ that is invariant under $\rho(G)$. So in this case you are asking for invariant points in the tree for $GL_2(K)$ under a compact subgroup $\rho(G)$ of $GL_2(K)$. There are various possibilities, and there is a nice paper of Bellaiche and Chenevier Sous-groupes de $GL_2$ et arbres discussing the possibilites.

When you say that you would like all integral models to be semi-simple, I'm not sure what you mean; but it say the represenation over $K$ is irreducible, then it will never happen that all the integral models have semi-simple reduction, if they are not themselves irreducible. (This is a proposition of Ribet, and is easily seen in the tree-based picture.)

Just to illustrate how one argues: if two points in the tree are $\rho(G)$-invariant, then so will be all the points on the line segment joining them. If $\rho(G)$ fixes an infinite half-line, this corresponds to an invariant one-dimensional space over $K$, and so means that $\rho$ is reducible over $K$. So if $\rho(G)$ is irreducible over $K$, then the fixed set of $\rho(G)$ will be bounded and convex, which already imposes some restrictions. (One an prove Ribet's result by these sort of considerations: basically, any extremal point of the fixed set of $\rho(G)$ will have non-simple reduction.)

Once you understand the $GL_2$ case you can imagine how to try and generalize things to the $GL_n$ case, although the combinatorics becomes more complictated, and (as Bellaiche and Chenevier note) it is harder to say anything nice, because when $n > 2$, $GL_n$ is the full automorphism group of the building.