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.
Edit: {The answer to your question, "...do we already know that a positive binary form represents arbitrarily long arithmetic progressions?" is yes. See the second paragraph below.}
If the relative density exists, so does the Dirichlet density and they are equal. The converse is not true in general. For primes in a given arithmetic progression, both densities exist. See Lang's Algebraic Number Theory, Ch. VIII.4 and XV. Given those facts, one approach to the problem would be trying to show that the relative density of the set of primes represented by a positive binary quadratic form actually exists (I have no idea how hard this might be).
On the other hand, if you only want to know about a.p.'s of primes represented by a positive quadratic form, a better approach might be answering the question, "Does Green-Tao still hold for sets of primes with positive Dirichlet density?" The answer is yes since G-T only requires that the limsup of the relative density be positive, and positive Dirichlet density implies positive limsup (if the limsup were 0, the lim would be 0).
Best Answer
Meyer (Über einen Satz von Dirichlet, Crelle 103 (1888)) proved that a primitive binary quadratic form with nonsquare discriminant represents infinitely many primes that lie in any given compatible residue class modulo a given integer N. He did so only for forms with even middle coefficient, and did not address the signs of primes, but both of these require only minor modifications.
This answer makes me wonder whether someone actually proved, with Dirichlet's methods, that a finite number of forms with coprime discriminants simultaneously represent infinitely many primes - by Chebotarev, this has to be true.