I am not an expert but the question:
"Does there exist a simply-connected closed Riemannian Ricci flat $n$-manifold with $SO(n)$-holonomy?"
is a well-known open problem. Note that Schwarzschild metric is a complete Ricci flat metric on $S^2\times\mathbb R^2$ with holonomy $SO(4)$, so the issue is to produce compact examples; I personally think there should be many. The difficulty is that it is hard to solve Einstein equation on compact manifolds. If memory serves me, Berger's book "Panorama of Riemannian geometry" discusses this matter extensively.
Bourguignon showed that if a compact manifold does not admit positive scalar curvature metrics, then any scalar flat metric (actually, any non-negative scalar curvature metric) is Ricci-flat; I suppose this is what you mean when you write "strongly scalar flat". But in three dimensions, the Ricci curvature determines the full curvature tensor, in particular, a Ricci-flat metric is flat. So any non-flat three-manifold which does not admit positive scalar curvature metrics will provide an example.
If I'm not mistaken, Gromov and Lawson proved that a compact three-manifold admits positive scalar curvature if and only if its prime decomposition contains no aspherical factors; note, this was before the Poincaré conjecture had been verified, so there would have been a caveat at the time of publication.
So $T^3\#S^1\times S^2$ is an example of a compact three-manifold of type (C). Note, $T^3\#S^1\times S^2$ is not flat as a non-trivial connected sum of compact manifolds of dimension at least three is never aspherical, but flat $n$-dimensional manifolds have universal cover $\mathbb{R}^n$.
In dimension four, you can sometimes use Seiberg-Witten invariants to rule out the existence of positive scalar curvature metrics, and then use the Hitchin-Thorpe inequality to rule out the existence of a Ricci-flat metric. For example, a compact Kähler surface with $b^+ \geq 2$ does not admit positive scalar curvature metrics; blowing up doesn't change this, but it eventually violates the Hitchin-Thorpe inequality.
A proof of Bourguignon's result can be found in Kazdan and Warner's paper Prescribing Curvatures, namely Lemma 5.2. As for the result of Gromov and Lawson, see Chapter IV, Theorem 6.18 of Lawson and Michelsohn's Spin Geometry and the discussion which follows.
Best Answer
If you don't care about completeness, here's a fairly simple way to construct such examples: Start with a compact Einstein manifold $(M^n,g)$ with Einstein constant $1$ (i.e., $\mathrm{Ric}(g) = (n{-}1)\,g$) that is not conformally flat. Now take the sine-cone, i.e., $\bigl(M\times(0,\pi),h\bigr)$ where $h = \mathrm{d}r^2 + (\sin r)^2\,g$. Then one easily computes that this is also an Einstein manifold with Einstein constant $1$, i.e., $\mathrm{Ric}(h) = n\,h$, but the Weyl curvature of $h$ (which is nonzero since $g$ is not conformally flat) blows up as $r$ approaches either $0$ or $\pi$.
(Also, if one just takes the ordinary cone, $\bigl(M\times(0,\infty),h\bigr)$ where $h = \mathrm{d}r^2 + r^2\,g$, then $h$ will be Ricci-flat, but the Weyl curvature of $h$ will blow up as $r\to 0$.)
Finally, as Anton Petrunin pointed out in the comment below, if $(M,g)$ is complete and Ricci-flat (i.e., Einstein with Einstein constant 0), but not conformally flat (equivalently, not flat), the Riemannian manifold $\bigl(M\times\mathbb{R}, h = \mathrm{d}r^2 + \mathrm{e}^{2r}\,g\bigr)$ will be a complete Einstein manifold with Einstein constant $-1$ whose Weyl curvature has unbounded norm as $r\to-\infty$.