I just wanted to remark that if $p$ is a prime such that
$\ell$ splits in $F = \mathbb{Q}(\sqrt{-p})$ for all $\ell \le N$,
then one may prove the existence of $N$ consecutive integers which
are norms of integers in $\mathcal{O}_F$, providing one is willing
to assume a standard hypothesis about prime numbers, namely,
Schinzel's Hypothesis H.
First, note the following:
Lemma 1: If $C$ is an abelian group of odd order, then there exists a finite
(ordered) set $S = \{c_i\}$ of elements of $C$ such that every element in $C$ can be
written in the form
$\displaystyle{\sum \epsilon_i \cdot c_i}$
where $\epsilon_i = \pm 1$.
Proof: If $C = A \oplus B$, take $S_C = S_A \cup S_B$. If $C
= \mathbb{Z}/n \mathbb{Z}$ then take $S = \{1,1,1,\ldots,1\}$ with
$|S| = 2n$.
Let $C$ be the class group of $F$. It has odd order, because
$2$ splits in $F$ and thus $\Delta_F = -p$.
Let $S$ be a set as in the lemma. Let $A$ denote an ordered
set of distinct primes $\{p_i\}$ which split in $\mathcal{O}_F$ such
that one can write $p_i = \mathfrak{p}_i \mathfrak{p}'_i$ with
$[\mathfrak{p}_i] = c_i \in C$, where $c_i$ denotes a set of elements
whose existence was shown in Lemma 1.
Lemma 2: If $n$ is the norm of some ideal $\mathfrak{n} \in \mathcal{O}_F$,
and $n$ is not divisible by any prime $p_i$ in $A$, then
$$n \cdot \prod_{A} p_i$$
is the norm of an algebraic integer in $\mathcal{O}_F$.
Proof: We may choose
$\epsilon_i = \pm 1$ such that
$\displaystyle{[\mathfrak{n}] + \sum \epsilon_i \cdot c_i = 0 \in C}$.
By assumption, $[\mathfrak{p}_i] = c_i \in C$ and thus
$[\mathfrak{p}'_i] = -c_i \in C$. Hence the ideal
$$\mathfrak{n}
\prod_{\epsilon_i = 1}
{ \mathfrak{p}}
\prod_{\epsilon_i = -1}
\mathfrak{p}'$$
is principal, and has the desired norm.
By the Chebotarev density theorem (applied to the Hilbert class field
of $F$), there exists a set $A$ of primes as above which avoids any fixed finite set of
primes. In particular, we may
find $N$ such sets which are pairwise distinct and which contain
no primes $\le N$. Denote these sets by $A_1, \ldots, A_N$.
By the Chinese remainder theorem, the set of integers $m$ such that
$$m \equiv 0 \mod p \cdot (N!)^2$$
$$m + j \equiv 0 \mod \prod_{p_i \in A_j} p_i, \qquad 1 \le j \le N$$
is of the form $m = d M + k$ where $0 \le k < M$, $d$ is arbitrary, and $M$
is the product of the moduli.
Lemma 3: Assuming Schinzel's Hypothesis H, there exists infinitely many integers
$d$ such that
$$ P_{dj}:= \frac{dM + k + j}{j \cdot \prod_{p_i \in A_j} p_i}$$
are simultaneouly prime for all $j = 1,\ldots,N$.
Proof: By construction, all these numbers are coprime to $M$ (easy check).
Hence, as $d$ varies, the greatest common divisor of the product of these
numbers is $1$, so Schinzel's Hypothesis H applies.
Let $\chi$ denote the quadratic character of $F$. Note that
$dM + k + j = j \mod p$, and so
$\chi(dM + k + j) = \chi(j) = 1$ (as all primes less than $N$ split in $F$).
Moreover, $\chi(p_i) = 1$ for all
primes $p_i$ in $A_j$
by construction. Hence
$\chi(P_{dj}) = 1$. In particular, if $P_{dj}$ is prime, then
$P_{dj}$ and $j \cdot P_{dj}$ are norms of (not necessarily principal) ideals in
the ring of integers of $F$. By Lemma 2, this implies that
$$dM + k + j = j \cdot P_{dj} \prod_{p_i \in A_j} p_i$$
is the norm of some element of $\mathcal{O}_F$ for all $j = 1,\ldots, N$.
One reason to think that current sieving technology will not be sufficient
to answer this problem is the
following: when Sieving produces a non-trivial lower bound, it usually
produces a pretty good lower bound. However, there are no good (lower) bounds known for the following problem: count the number of integers $n$
such that $n$, $n+1$, and $n+2$ are all sums of two squares. Even for the
problem of estimating the number of $n$ such that $n$ and $n+1$ are both sums of squares is tricky - Hooley implies that the natural sieve does not give lower bounds
(for reasons analogous to the parity problem). Instead,
he relates the problem to sums of the form
$\displaystyle{\sum_{n < x} a_n a_{n+1}}$ where
$\sum a_n q^n = \theta^2$ is a modular form. In particular, he
implicitly uses
automorphic methods which won't work
with three or more terms.
Best Answer
The answer to both your questions is positive and indeed, every given pattern of quadratic residues and non-residues of fixed length appears among consecutive elements of ${\mathbb F}_p$, for all $p$ large enough; moreover, it appears about the expected number of times. This is non-trivial, but fairly standard.
Fix $\epsilon=(\epsilon_1,\ldots,\epsilon_k)\in\{-1,1\}^k$ ("the pattern") and for a prime $p$, let $N_\epsilon(p)$ denote the number of those $x\in{\mathbb F}_p$ with $$ \left(\frac{x+1}p\right)=\epsilon_1,\ldots, \left(\frac{x+k}p\right)=\epsilon_k, $$ where $\left(\frac xp\right)$ is the Legendre symbol. Clearly, we have \begin{align*} N_\epsilon(p) &= 2^{-k}\sum_{x=0}^{p-k-1} \prod_{j=1}^k \left(1+\epsilon_j\left(\frac{x+j}{p}\right) \right) \\ &= 2^{-k}\sum_{x=0}^{p-1} \prod_{j=1}^k \left(1+\epsilon_j\left(\frac{x+j}{p}\right) \right) - \theta \frac k2,\quad 0\le \theta\le 1, \end{align*} as the contribution of each $x\in[p-k,p-1]$ to the whole sum is at most $2^{k-1}$.
Expanding the product, one can write $N_\epsilon(p)$ as a sum of the main term $2^{-k}\sum_x1=2^{-k}p$ and $2^k-1$ remainder terms, each of the form $2^{-k}\sum_x \left(\frac{Q(x)}p\right)$ with a non-square polynomial $Q(x)$ of degree at most $k$. Now, Weil's bound implies that each of these remainder terms does not exceed $(k-1)\sqrt p$ in absolute value; as a result, $$ N_\epsilon(p)=2^{-k}p+\theta k(\sqrt p+1/2),\ |\theta|<1, $$ which is certainly positive for $p$ sufficiently large (like $p>\exp(ck)$ with a suitable constant $c$).
This argument readily extends to count, say, the number of those elements $x$ of a finite field such that for a given system of square-free, pairwise co-prime polynomials $P_1,\ldots,P_k\in{\mathbb Z}[X]$, the values $P_1(x),\ldots,P_k(x)$ follow a prescribed quadratic residue / non-residue pattern. Indeed, in a similar way one can handle the joint distribution of $P_1(x),\ldots P_k(x)$ in the cosets of any subgroup of the multiplicative group of a finite field, not just the subgroup of quadratic residues.