The answer was provided to me by Eric Cator. Here is his formulation which is almost the same.
Let $Z$ be a Gaussian random vector $\mathcal{N}(0,I_n)$ for some positive integer $n$. Let $q>0$, $F\subset\mathbb{R}^n$ be a cone, and let $F^o$ be its polar cone. Then,
$$\mathbb{P}\left(\|Z+\mu\|>q|Z+\mu\in F\right)\leq \mathbb{P}\left(\|Z\|>q|Z\in F\right), \quad \forall \mu\in F^o.$$
Proof:
We rewrite the conditional probability in polar coordinates. Let $S=\{\omega\in\mathbb{R}^n| \; \|\omega\|=1\}$ be the unit sphere in $\mathbb{R}^n$. Denote $\nu$ the surface measure on $S$. We have
$$\mathbb{P}\left(\|Z+\mu\|>q|Z+\mu\in F\right) = \frac{\int_{r=q}^{\infty}\int_{\omega\in S \cap F}t^{n-1}e^{-\frac{1}{2}\|tw-\mu\|^2}\nu(d\omega)dt}{\int_{r=0}^{\infty}\int_{\omega\in S \cap F}s^{n-1}e^{-\frac{1}{2}\|sw-\mu\|^2}\nu(d\omega)ds} $$
This means that the conditional density of $\|Z+\mu\|$ provided that $\{Z+\mu\in F\}$ is given by
$$g_{\mu}(t) = \frac{t^{n-1}e^{-\frac{1}{2}t^2}\int_{\omega\in S \cap F}e^{tw^T\mu}\nu(d\omega)}{\int_{r=0}^{\infty}\int_{\omega\in S \cap F}s^{n-1}e^{-\frac{1}{2}s^2+sw^T\mu}\nu(d\omega)ds}.$$
Denote $A(\mu)$ the normalization constant in the previous display, that is
$$A(\mu) = \int_{r=0}^{\infty}\int_{\omega\in S \cap F}s^{n-1}e^{-\frac{1}{2}s^2+sw^T\mu}\nu(d\omega)ds.$$
Define
$$G_{\mu}(t) = \frac{g_0(t)}{g_{\mu}(t)} = \frac{A(\mu)}{A(0)}\frac{\int_{\omega\in S \cap F}\nu(d\omega)}{\int_{\omega\in S \cap F}e^{t\omega^T\mu}\nu(d\omega)}.$$
Since $\mu\in F^o$, then
$$\forall\omega\in S \cap F, \quad \omega^T\mu \leq 0.$$
Thus, function $t\mapsto G_{\mu}(t)$ is nondecreasing over $(0,\infty)$ whatever the value of $\mu\in F^o$.
Let $h$ be some measurable nondecreasing function defined on $\mathbb{R}_+$. Note that for any couple of nonnegative real numbers $(r_1,r_2)$, since both $G_{\mu}(t)$ and $h(t)$ are nondecreasing functions in $t$ over $(0,\infty)$, we have
\begin{equation}
(h(r_1)-h(r_2))(G_{\mu}(r_1)-G_{\mu}(r_2))\geq 0
\qquad (1)
\end{equation}
Let $R_1$ and $R_2$ be two i.i.d. random variables with a common density defined on $(0,\infty)$. We have, due to (1),
$$\mathbb{E}\left[\left(h(R_1)-h(R_2)\right)\left(G(R_1)-G(R_2)\right)\right]\geq 0.$$
We deduce that
\begin{equation}
\mathbb{E}\left[h(R_1)G_{\mu}(R_1)\right]\geq \mathbb{E}\left[h(R_1)\right]\mathbb{E}\left[G_{\mu}(R_1)\right].
\qquad (2)
\end{equation}
Assume now that $R_1$ has the density $g_{\mu}$, and denote $\mathbb{E}_{\mu}$ (resp. $\mathbb{E}_0$) the expectation under $g_{\mu}$ (resp. $g_0$). We have now
\begin{align*}
\mathbb{E}_{\mu}\left[h(R_1)G_{\mu}(R_1)\right] & = \mathbb{E}_{0}\left[h(R_1)\right] \\
\mathbb{E}_{\mu}\left[G_{\mu}(R)\right] & = 1.
\end{align*}
The second line in the previous display comes from the fact that $g_0$ is a density over $(0,\infty)$. Hence, due to (2), we may write
\begin{equation}
\mathbb{E}_{\mu}\left[h(R)\right] \leq \mathbb{E}_{0}\left[h(R)\right].
\qquad (3)
\end{equation}
Set $h(t) = 1_{t>q}$ (the indicator function of the set $(q,\infty)$). Function $h$ is nondecreasing over $(0,\infty)$, therefore we could apply (3) on it and get
$$\frac{\int_{r=q}^{\infty}\int_{\omega\in S \cap F}t^{n-1}e^{-\frac{1}{2}\|tw^T\mu\|^2}\nu(d\omega)dt}{\int_{r=0}^{\infty}\int_{\omega\in S \cap F}s^{n-1}e^{-\frac{1}{2}\|sw^T\mu\|^2}\nu(d\omega)ds} \leq \frac{\int_{r=q}^{\infty}\int_{\omega\in S \cap F}t^{n-1}\nu(d\omega)dt}{\int_{r=0}^{\infty}\int_{\omega\in S \cap F}s^{n-1}\nu(d\omega)ds}. $$
This is exactly what we claim.
