Having scalar product above $\rho$ is equivalent to having spherical distance above $2\theta$, where $\theta=\cos(\rho)/2$. I consider first the case of the whole sphere, and I'll discuss the positive orthant at the end.

### Upper bound

Let us begin by finding an upper bound. Suppose that you have points $x_1,\ldots,x_N$ on the surface of the $(d-1)$-sphere, at spherical distance at least $2\theta$ from each other. Then there are disjoint open cones $C_i$ in the unit $d$-sphere, where $C_i$ is the interior of the convex hull of
- the vertex $0$,
- the points $y$ on the $(d-1)$-sphere with $\langle x_i,y\rangle=\sin(\theta)$

The cones $C_i$ and $C_j$ are disjoint whenever $i \neq j$, because points in $C_i$ have angles with $x_i$ of less than $\theta$, and points in $C_j$ have angles with $x_j$ of less than $\theta$, but the angle between $x_i$ and $x_j$ is at least $2\theta$.

This means that $N\cdot|C_1|_d\leq |\mathbb B^d|_d$, where $|\cdot|_d$ denotes the Lebesgue volume in dimension $d$ and $\mathbb B^d$ is the unit ball in $\mathbb R^d$. Since the height of the cones is $\cos(\theta)$, their volume is
$$|C_1|_d=\frac1d|\mathbb B^{d-1}|_{d-1}\cdot\sin(\theta)^{d-1}\cdot\cos(\theta).$$
and therefore
$$N \leq \frac{|\mathbb B^d|_d}{|\mathbb B^{d-1}|_{d-1}}\cdot\frac{d}{\cos(\theta)}\cdot\sin(\theta)^{1-d}.$$
Since the first ratio goes to zero (as $1/\sqrt d$, up to constant), $N$ has to be asymptotically less than $\lambda^d$ for all $\lambda>1/\sin(\theta)$.

### Lower bound

Let $x_1,\ldots,x_N$ be a maximal family of points on the sphere given the constraint that two points are at spherical distance at least $2\theta$. The sphere is covered by the closed caps centred at the $x_i$ with spherical radius $2\theta$, since any point not covered could be added to the maximal family. Then the unit ball is covered by closed cones $C'_i$, where $C'_i$ is the set of $ty$ with
- $0\leq t\leq 1/\cos(2\theta)$,
- $y$ is a point on the unit sphere with $\langle x_i,y\rangle\leq\sin(2\theta)$.

In other words, the base of the cone $C'_i$ is tangent to the unit sphere, and is just large enough so that it contains the spherical cap associated to $x_i$. This way, the $C'_i$ cover the sphere, so they have to cover the unit ball also.

This means that, considering the Lebesgue measure,
$$ |\mathbb B^d|_d\leq N\cdot\frac1d|\mathbb B^{d-1}|_{d-1}\tan(2\theta)^{d-1}\cdot1 $$
and $N$ satisfies the bound
$$ d\frac{|\mathbb B^d|_d}{|\mathbb B^{d-1}|_{d-1}}\tan(2\theta)^{1-d}\leq N. $$

### The positive orthant

Writing $K_d$ for the ratio $|\mathbb B^d|_d/|\mathbb B^{d-1}|_{d-1}$, the above shows that the maximal number of points $N_\text{max}$ at spherical distance at least $2\theta$ on the surface of $\mathbb S^{d-1}$ satisfies
$$ K_d d\tan(2\theta)^{1-d}\leq N_\text{max}\leq K_d d\frac1{\cos(\theta)}\sin(\theta)^{1-d}. $$
In other words, it grows exponentially in $d$ ($\ln(N)/d$ is bounded above and below by constants depending on $\theta$ but not $d$). For this we need the fact that $K_d$ decreases as $d^{-d/2}$, which one can show using explicit expression for the [volume of a $d$-dimensional ball](https://en.wikipedia.org/wiki/Volume_of_an_n-ball) together with [Stirling's formula](https://en.wikipedia.org/wiki/Stirling%27s_approximation).

Of course the upper bound still holds on the positive orthant. Using an averaging argument (see below), we can show that for the portion of sphere you consider,
$$ 2K_d d(\tan(2\theta)/2)^{1-d}\leq\widetilde N_\text{max}\leq K_d d\frac1{\cos(\theta)}\sin(\theta)^{1-d}, $$
so the constant in the exponent is asymptotically $1/\theta$:
$$ \limsup_{\theta\to0}\limsup_d\left|\frac1d\ln(\widetilde N_\text{max})+\ln(\theta)\right| = 0. $$

### Averaging argument

(Note: the averaging argument is not really required here, the proof would work directly with a fraction of the sphere, but it's a nice argument.)

Let $X=\{x_1,\ldots,x_N\}$ be a set of points on $\mathbb S^{d-1}$ at spherical distance at least $2\theta$. We can divide $\mathbb R^d$ in $2^d$ open orthants $S_1,\ldots,S_{2^d}$ (and a set of measure zero that will not contribute in the following). For each of them, there exists a rotation $R_i$ sending it to $S_1$, i.e. $S_1 = R_iS_i$. Let $R$ be a random rotation in $SO_d(\mathbb R)$, distributed according to the Haar measure (say left-invariant, although the left- and right-invariant measures are actually the same in this case), and $\mathbb E$ the corresponding expectation. Then
$$ \begin{align*}
N &= \mathbb E[\#RX] \\
  &= \mathbb E[\#(S_1\cap RX)] + \cdots + \mathbb E[\#(R_{2^d}^{-1}S_1\cap RX)] \\
  &= \mathbb E[\#(S_1\cap RX)] + \cdots + \mathbb E[\#(S_1\cap R_{2^d}RX)] \\
  &= 2^d\mathbb E[\#(S_1\cap RX)].
\end{align*} $$
It means that there must exist an $R_0$ with $\#(S_1\cap R_0X)\geq N/2^d$, otherwise the expectation could not be this high. Now the set $S_1\cap R_0X$ consists of at least $N/2^d$ points in a fixed orthant at spherical distance at least $2\theta$. Using this reasoning with the $N_\text{max}$ described in the previous section gives the expected estimate on $\widetilde N_\text{max}$.