Here is an elaboration of my comment to the question. Suppose that the primes are (pre)-periodic $\pmod k$ (with $k>2$). Let $\chi$ be a non-trivial character $\pmod k$. Then the assumption on the primes implies that
$$
S(x,\chi) = \sum_{p\le x} \chi(p),
$$
is bounded.
Now consider
$$
\log L(s,\chi) = \sum_{j=1}^{\infty} \frac{1}{j} \sum_{p} \frac{\chi(p)^j}{p^{js}}.
$$
A priori this is well defined and analytic in Re$(s)>1$. The terms with $j\ge 2$ are analytic in Re$(s)>1/2$. The assumption on the periodicity of the primes gives that
$$
\sum_{p} \frac{\chi(p)}{p^s} = \int_{1}^{\infty} S(x,\chi) \frac{s}{x^{s+1}} dx,
$$
is analytic in Re$(s)>0$. Thus we conclude that $\log L(s,\chi)$ is analytic in Re$(s)>1/2$ (this is the Riemann Hypothesis, but of course a false assumption implies many things!).
Now suppose that $\chi$ is a non-trivial quadratic character. Then from above we see that for Re$(s)>1/2$ we have
$$
\log L(s,\chi) = O(1)+ \frac{1}{2} \sum_{p} \frac{\chi(p)^2}{p^{2s}} = O(1) + \frac{1}{2} \sum_{p} \frac{1}{p^{2s}},
$$
since $\chi$ is quadratic.
If we let $s$ take real values tending to $1/2$ from above, this gives a contradiction (as $L(1/2,\chi)$ is bounded, whereas exponentiating the above would imply that it goes to infinity).
Alternatively take $\chi$ to be a complex character (this works for $k\neq 3, 4, 6, 8$) and then the prime square terms above are also bounded (for Re$(s)>0$) and we conclude that $\log L(s,\chi)$ is analytic in Re$(s)>1/3$. Hence the $L$-function is non-zero in that region, and by the functional equation (say $\chi$ is primitive) we conclude that the $L$-function can only have trivial zeros. This is a contradiction.
Thirdly, as noted before Daniel Shiu's result on Strings of Congruent primes (JLMS 2000) also gives the result (although this is harder than the proof above). Nowadays the Maynard(-Tao) machinery gives relatively easy access to such results (and in a stronger form).
Finally, as Terry Tao observes in the comments, Gowers's question on why the Riemann zeta function has zeros is certainly relevant here, and see also my answer to that question. In that spirit, one may say that the primes are not periodic $\pmod k$, because the integers are!
You asked for a heuristic answer.
There is an heuristic argument that infinitely many such partial sums should exist. Consider $P(k)$, an heuristic estimate of the probability that the partial sum of the first $k+1$ primes would be divisible by $p_k$. Now $$p_k \sim k \log k$$ and if only random chance were involved, $$P(k) \approx \frac1{p_k} \sim \frac1{k \log k}$$
In that case, the expected number of primes with the property you want would be something like
$$\int_2^\infty \frac1{x \log x}\,dx$$
and that integral diverges to infinity.
The reason it seems so rare is that the rate of divergence is like $\log(\log x)$ and while that function goes to infinity, "nobody ever sees it do so."
On the other hand, proving that there an infinite number of such values of $k$ (in the same sense that Euclid's argument proves there is no last prime) is probably quite difficult. And if the conjecture that there are an infinite number of such values of $k$ turned out to be false, proving that some particular $k$ is the last one with this property would seem to be even harder.
Best Answer
I wonder if for questions of this type, the following probabilistic paradigm is useful:
In other words, what can we say about the distribution of primes in the sequence of sums
$$S_{\xi}(n)=\sum_{i=1}^n \xi_i p_i,$$
where $p_i$ is the $i$th prime and $\{\xi_i\}$ is a sequence of i.i.d. Bernoulli random variables? The asymptotic needs to be a.s. with respect to the Bernoulli measure.