Your observation is true, and a somewhat stronger result is true: you will almost always be able to find a smaller prime which divides the chosen prime and leaves a remainder of either $4$ or $6$. In the rare instances where a remainder of either $4$ or $6$ cannot be found (such as for $31$), there can be found a remainder of $12$, which is a multiple of both $4$ and $6$.
Consider the set of positive integers greater than $1$ having no factor of $2$ or $3$. They are represented as having the form $6k\pm 1$ for $k\ge 1$. Note that the smallest member of this set is the prime number $5$, and all prime numbers other than $2,3$ are included in this set. Note further that this set is closed under multiplication: the product of any two members of the set is another member of the set (i.e. a nmumber lacking any factor of $2$ or $3$), and if a member of the set has factors, its factors are members of the set also (i.e. numbers lacking any factor of $2$ or $3$).
Case A: If you have a prime number of the form $p_k=6k-1 \ge 11$, you can subtract $4$ from it to get a smaller number $6m+1$ where $m=k-1$. Either $6m+1$ is a prime, or it has prime factors. If $6m+1=p_m$ is prime, then $p_k=p_m+4$, and dividing $p_k$ by $p_m$ leaves a remainder of $4$. If $6m+1$ is not prime, then it has at least one smaller prime factor: $6m+1=n\cdot p_m$ and $p_k=n\cdot p_m+4$. But $p_m \ge 5>4$, so here again dividing $p_k$ by $p_m$ leaves a remainder of $4$.
Case B: If you have a prime number of the form $p_k=6k+1 \ge 13$, you will almost always be able to subtract $6$ from it to get a smaller number $6m+1$ where $m=k-1$. Either $6m+1$ is a prime, or it has prime factors. If $6m+1=p_m$ is prime, then $p_m \ge 7$ and $p_k=p_m+6$. Thus dividing $p_k$ by $p_m$ leaves a remainder of $6$. If $6m+1$ is not prime, then it has at least one prime factor: $6m+1=np_m$ and $p_k=np_m+6$. If $p_m \ge 7>6$, then dividing $p_k$ by $p_m$ always leaves a remainder of $6$.
Case C: The rare case for $p_k=6k+1 \ge 13$ (as in Case B), but $6m+1$ only contains factors of $5$, such that $6m+1=5^r$ and $p_k=5^r+6$. Here, dividing $p_k$ by $5$ leaves a remainder of $1$. First, note that a number of the form $6m+1=5^r$ can only be an even power of $5$ if $r$ is even, not odd, i.e. $6m+1=5^{2s}=25^s$. This means $p_k=25^s+6$ which implies that $p_k-12=25^s-6=6n+1$. Either $6n+1$ is a prime, or it has prime factors. If we divide $p_k$ by $6n+1$ if it is a prime, or by any of its prime factors if it is not prime, there will be a remainder of $12$, provided that neither $6n+1$ nor all of its prime factors are smaller than $12$. We know that $6n+1$ has no factors of $2$ or $3$, and we can see immediately that $6n+1=25^s-6$ is not evenly divisible by $5$. I will prove that $25^s-6$ is not evenly divisible by $7$ or $11$, which means that any prime factor of $6n+1$ is larger than $12$, so dividing $p_k=(6n+1)+12$ by any of those prime factors will in fact leave a remainder of $12$.
Proof that $25^s-6$ is not evenly divisible by $7$ or $11$: By modular arithmetic, $25^s-6 \equiv 4^s+1 \bmod 7$. As $s$ increases from $1$, the values of $4^s+1 \bmod 7$ cycle: $5,3,2,5,3,2,\dots$. Since division of $25^s-6$ by $7$ always leaves a non-zero remainder, it is never divisible by $7$, and hence has no factors of $7$. Similarly, $25^s-6 \equiv 3^s+5 \bmod 11$. As $s$ increases from $1$, the values of $3^s+5 \bmod 7$ cycle: $8,3,10,9,6,8,3,10,9,6,\dots$. Since division of $25^s-6$ by $11$ always leaves a non-zero remainder, it is never divisible by $11$, and hence has no factors of $11$.
In summary, primes $p_k>11$ of the form $6k-1$ will always admit of a smaller prime that divides $p_k$ leaving a remainder of $4$. Primes $p_k>11$ of the form $6k+1$ will almost always admit of a smaller prime that divides $p_k$ leaving a remainder of $6$, and in the rare cases where that is not true, primes $p_k>11$ of the form $6k+1$ will then admit of a smaller prime that divides $p_k$ leaving a remainder of $12$, which is a multiple of both $4$ and $6$.
Best Answer
The Wikipedia article Dirichlet's theorem on arithmetic progressions answers your question:
In other words, for any $k$ with $0 < k < n$, the proportion of integers $i$ such that $r[i] = k$ is equal to $1/(n-1)$ (since $n$ is prime).
More formally: Given an integer $N > 0$, let $p_N(k)$ be the proportion of integers $i$ with $0 < i \le N$ such that $r[i] = k$. Then $p_N(k)$ tends to $1/(n-1)$ as $N$ tends to $\infty$.
It is possible to make a more precise statement concerning the rate of convergence of $p_N(k)$, but I don't expect that any more concrete result is possible.