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.
Here is a proof of this fact.
We start with a standard
Lemma 1. Any prime divisor $q$ of $1+x+x^2$ for an integer $x$ is either equal to 3 or is congruent to 1 modulo $3$.
Proof. If, on the contrary, that $q=3k+2$, then $x^3\equiv 1 \pmod q$ and also by Fermat's little theorem $x^{3k+1}\equiv 1 \pmod q$, therefore $x=x\cdot 1^k\equiv x (x^3)^k=x^{3k+1}\equiv 1\pmod q$, but then $1+x+x^2\equiv 3\pmod q$, a contradiction.
Now assume that an odd number $m=p^ap_1^2\ldots p_n^2$ satisfies $$m=\frac12\sigma(m)=\frac{1+p+\ldots+p^a}2\cdot \prod_{i=1}^n (1+p_i+p_i^2).\tag{1}\label{1}$$
If $a$ is even, than RHS of \eqref{1} is not integer. So, $a$ is odd and $1+p+\ldots+p^a=(p+1)(1+p^2+p^4+\ldots+p^{a-1})$, thus $(p+1)/2$ divides $m$.
Lemma 2. 3 divides $m$.
Proof.
Since $p$ and $(p+1)/2$ are coprime, some $p_i$ must divide $(p+1)/2$, let it be $p_1$. We have $p\geqslant 2p_1-1$, thus $p^2\geqslant (2p_1-1)^2>1+p_1+p_1^2$. Next, if $1+p_1+p_1^2=p$, then $(p+1)/2=1+p_1(p_1+1)/2$ is not divisible by $p_1$, a contradiction. Therefore, $y:=1+p_1+p_1^2$ is not equal to $p$ and is less than $p^2$. Then $y$ has a prime divisor different from $p$, and by \eqref{1} it is some $p_i$, let it be $p_2$. By Lemma 1, $p_2$ is either equal to 3 or congruent to $1$ modulo 3. In the second case 3 divides $1+p_2+p_2^2$, which in turn divides $m$ by \eqref{1}. Lemma 2 is proved.
Since $(p+1)/2$ divides $m$, it is odd, thus $p\ne 3$. Therefore some $p_i$ is equal to 3. Thus, by \eqref{1}, 1+3+3^2=13 divides $m$.
If $p=13$, then $(p+1)/2=7$ divides $m$, then so does $1+7+7^2=57=3\cdot 19$, and so does $1+19+19^2=381=3\cdot 127$, and $m$ is already divisible by $(1+7+7^2)(1+19+19^2)(1+127+127^2)$, so 27 divides $m$, a contradiction.
If $p\ne 13$, then some $p_i$ is 13, and $m$ is divisible by $1+13+13^2=183=3\cdot 61$. If $p=61$, then $31=(p+1)/2$ divides $m$, so does $1+31+31^2=3\cdot 331$, and 27 divides $(1+13+13^2)(1+31+31^2)(1+331+331^2)$ which divides $m$, a contradiction. If some $p_i$ is 61, then $1+61+61^2=3\cdot 13\cdot 97$ divides $m$. If $p=97$, then $49=(p+1)/2$ divides $m$, and we again conclude that $m$ is divisible by $27$ (because of divisors 7,13,61). Otherwise, some $p_i$ is equal to 97, and we make the same conclusion.
Best Answer
It is very likely that there are only finitely many primes with this property. Heuristically, the probability of this happening for the $N$th prime is $2^{2-N}$, and $\sum_{N \geq 2} 2^{2-N} =2$, so we expect only a couple primes where this happens.
I can't see how one would ever hope to prove this. However, we can prove there are not many such primes in some large intervals.
If $p$ and $q$ are two such primes with $p<q$, then $\left( \frac{ \ell}{pq}\right)$ is either the constant function $1$ for $\ell<p$ or the constant function $-1$. Thus $\left( \frac{m}{pq} \right)$ is either the constant function $1$ for $m < pq$ or the Möbius function. The second case is a Siegel zero-type phenomenon, but the first case is easier to rule out.
Since most numbers $< p^{\sqrt{e}}$ have only prime factors $<p$, the average of the Legendre symbol $\left( \frac{m}{pq} \right)$ over numbers $< p^{\sqrt{e}}$ is positive and large in the first case - there are more $1$s then $-1$s. By Burgess, this is impossible unless $pq > (p^{\sqrt{e}})^{4-\epsilon}$, or $q> p^{ 4 \sqrt{e}-1 -o(1)}$.
So every perfectl equidistributed prime between $p$ and $ p^{ 4 \sqrt{e}-1 -o(1)}$ must have the opposite sign (of the Legendre symbol applied to 2) from $p$. Since these perfectly equidistributed primes must also have opposite signs from each other, there can be at most one more perfectly equidistributed prime in that range.
This means the number of perfectly equidistributed primes $<n$ is at most $$ \left(\frac{2}{ \log (4 \sqrt{e}-1)} + o(1)\right) \log \log n.$$
So there can be very few such primes in a given range.