Here's a heuristic that suggests why arbitrarily large strings of consecutive numbers should be representable by some binary quadratic form. For simplicity consider a prime $p$ that is $3\pmod 4$ so that $-p$ is a fundamental discriminant. Suppose that $p$ has been chosen in such a way that all the primes $\le k$ are quadratic residues $\pmod p$. Suppose there are $h$ quadratic forms of discriminant $-p$ (and note that there is only one genus).

Recall that a number $n$ is represented by some form of discriminant $-p$ if every prime factor $\ell$ of $n$ satisfies $\chi(\ell)=1$. As discussed in my answer to Achieving consecutive integers as norms from a quadratic field we should expect to find many strings of $k$ consecutive numbers, each of which is representable by some form of discriminant $-p$.

In my answer to that question, I focused on such strings of (almost) prime numbers, but the same Hardy-Littlewood heuristics would predict lots of strings $n+1$, $\ldots$, $n+k$ where each $n+j$ is divisible only by primes that are quadratic residues $\pmod p$, and each $n+j$ has a typical number of prime factors. Under the restriction that all prime factors of $n$ are quadratic residues $\pmod p$, if $n$ is large then typically it will have about $\frac 12 \log \log n$ such prime factors. Moreover we may expect these prime factors to be roughly equally distributed in the class group (which is of fixed size $h$). Thus since there are $2^{\omega(n)}$ factorizations of $n$ as a product of two ideals, we would expect that typically there are many such factorizations with one of the factors lying in a prescribed ideal class.

Summarizing one would expect that there are many strings $n+j$ ($1\le j\le k$) with each $n+j$ composed of about $\frac 12\log \log n$ primes that are all quadratic residues $\pmod p$, and (typically) each such $n+j$ would be represented by every form of discriminant $-p$.

It should be possible with a little effort to turn this into a precise Hardy-Littlewood type conjecture, but I don't see any hope of a proof.

The heuristic described above is probably classical. One place where this heuristic is described is a paper of Blomer and Granville: see pages 9 and 10 of http://www.dms.umontreal.ca/~andrew/PDF/quadraticforms.pdf

Here's a heuristic that suggests why arbitrarily large strings of consecutive numbers should be representable by some binary quadratic form. For simplicity consider a prime $p$ that is $3\pmod 4$ so that $-p$ is a fundamental discriminant. Suppose that $p$ has been chosen in such a way that all the primes $\le k$ are quadratic residues $\pmod p$. Suppose there are $h$ quadratic forms of discriminant $-p$ (and note that there is only one genus).

Recall that a number $n$ is represented by some form of discriminant $-p$ if every prime factor $\ell$ of $n$ satisfies $\chi(\ell)=1$. As discussed in my answer to Achieving consecutive integers as norms from a quadratic field we should expect to find many strings of $k$ consecutive numbers, each of which is representable by some form of discriminant $-p$.

In my answer to that question, I focused on such strings of (almost) prime numbers, but the same Hardy-Littlewood heuristics would predict lots of strings $n+1$, $\ldots$, $n+k$ where each $n+j$ is divisible only by primes that are quadratic residues $\pmod p$, and each $n+j$ has a typical number of prime factors. Under the restriction that all prime factors of $n$ are quadratic residues $\pmod p$, if $n$ is large then typically it will have about $\frac 12 \log \log n$ such prime factors. Moreover we may expect these prime factors to be roughly equally distributed in the class group (which is of fixed size $h$). Thus since there are $2^{\omega(n)}$ factorizations of $n$ as a product of two ideals, we would expect that typically there are many such factorizations with one of the factors lying in a prescribed ideal class.

Summarizing one would expect that there are many strings $n+j$ ($1\le j\le k$) with each $n+j$ composed of about $\frac 12\log \log n$ primes that are all quadratic residues $\pmod p$, and (typically) each such $n+j$ would be represented by every form of discriminant $-p$.

It should be possible with a little effort to turn this into a precise Hardy-Littlewood type conjecture, but I don't see any hope of a proof.

The heuristic described above is probably classical. One place where this heuristic is described is a paper of Blomer and Granville: see pages 9 and 10 of http://www.dms.umontreal.ca/~andrew/PDF/quadraticforms.pdf