This reference contains the best result of this kind currently known for $\mu(n)$:
Tadej Kotnik and Herman te Riele The Mertens Conjecture revisited. Algorithmic number theory, 156--167, Lecture Notes in Comput. Sci., 4076, Springer, Berlin, 2006.
They prove that $\limsup_{x \rightarrow +\infty}M(x)/\sqrt{x} \geq 1.218$ and that $\liminf_{x \rightarrow +\infty}M(x)/\sqrt{x} \leq -1.229$. Here
$
M(x) = \sum_{n \leq x}\mu(n)
$
is the conventional notation for the summatory function of the Möbius function. Their proof is a mixture of analytic number theory and large scale computations. They also have a survey of what is known and what is conjectured about the size of $M(x)$.
Now on to the Liouville function $\lambda(n)$ and its summatory function $L(x)$. The latter is very closely connected with $M(x)$, for
$
\sum_{n \leq \sqrt{x}}\mu(n)L\left(\frac{x}{n^2}\right) = \sum_{n \leq \sqrt{x}}\mu(n)\sum_{m \leq x/n^2}\lambda(m) = \sum_{N \leq x}\sum_{mn^2 = N}\mu(n)\lambda(m) = \sum_{N \leq x}\mu(N) = M(x).
$
Thus
$
|M(x)| \leq \sum_{n \leq \sqrt{x}}|L\left(\frac{x}{n^2}\right)|
$
and so the assumption
$
L(x) = o\left(\frac{\sqrt{x}}{\log^{1 + \epsilon}(x)}\right)
$
for example, leads to a contradiction with the Kotnik-te Riele result (or earlier results) for any $\epsilon > 0$.
My guess is that if one looks up the old (pre-computer) results on $|M(x)|$ from below, one might prove that $\limsup_{x \rightarrow +\infty}|L(x)|/\sqrt{x} > 0$. This may even be in the literature somewhere.
Alternatively
$
\sum_{n = 1}^{\infty}\lambda(n)n^{-s} = \prod_{p}\sum_{k=0}^{\infty}\lambda(p^k)p^{-ks} = \prod_{p}\sum_{k=0}^{\infty}(-1)^kp^{-ks} = \prod_{p}(1 + p^{-s})^{-1} = \frac{\zeta(2s)}{\zeta(s)}
$
by the Euler product formula, and from here on it is more elementary than it was with $M(x)$ in the argument that David Speyer gave, because we don't need the zeros on the critical line. For $\zeta(2s)$ has a pole at $s = 1/2$ which is not canceled by a pole of $\zeta(s)$ at the same point. Thus $L(x) = O(x^{\alpha})$ is impossible with $\alpha < 1/2$ by partial summation.
For multiplicative functions of modulus $1$, the situation is much less clear. For simplicity, assume that $f(n)$ is a totally multiplicative function (multiplicative and $f(p^k) = f(p)^k$) with $|c_p| = 1$ where $c_p = f(p)$. The Liouville function is the case
$c_p \equiv -1$. Then
$
\sum_{n = 1}^{\infty}f(n)n^{-s} = \prod_{p}\frac{1}{1 - c_pp^{-s}}{\quad},{\quad}A(x) = \sum_{n \leq x}f(n)
$
by the Euler product formula. The basic principle is that if $A(x) = O(x^{\alpha})$, then the Dirichlet series on the left hand side is convergent in the half plane $\sigma > \alpha$, so the sum is holomorphic in that half plane. If we can find a singularity $s_0$ of the product on the right hand side with $\mathrm{Re}(s_0) = \sigma_0$, that tells us that
$A(x) = O(x^{\alpha})$ with $\alpha < \sigma_0$ is impossible. The bad thing now is that the product may diverge at a point without having a singularity there, because the product may diverge to zero, and a holomorphic function may be zero at a point without being singular there.
But it is straightforward to show that if $\mathrm{Re}(c_p) \geq \delta > 0$ for all $p$, then $A(x) = O(x^{\alpha})$ is impossible for any $\alpha < 1$, by showing that the series
$
\sum_{p}c_pp^{-\sigma}
$
goes to infinity as $\sigma \rightarrow 1^{+}$.
This question is extremely close to this one
Covering the primes by 3-term APs ?
though not exactly the same.
For much the same reasons as described in the answer given there, the answer to your question is almost certainly yes, but a proof is beyond current technology, exactly as you suggest. I'm not aware that the problem has a specific name.
To show that 3 belongs to a 3PAP is of course trivial: it belongs to 3,5,7 or 3,7,11. Showing that there are infinitely many such 3PAPs is, as you point out, a problem of the same level as difficulty as the Sophie Germain primes conjecture or the twin primes conjecture.
For a general p, I find it extremely unlikely that you could show that there is a k > 0 such that p + k, p + 2k are both prime without showing that there are infinitely many. Proving this for even one value of p would be a huge advance.
I think you could show that almost all primes p do have this property using the Hardy-Littlewood circle method.
Best Answer
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.