TL;DR
As stated the desired bound cannot hold. However, the bound holds when $v > 0$
- for $d \leq 3$;
- for $d > 3$, for a constant $C$ depending on $v$ that blows up as $v \to 0$;
- for any $d$, for a uniform constant $C$ independent of $v$ if we assume additional cut-offs in $r$.
For convenience I will prove the claims for $d = 2k+1$ odd, with $k \geq 1$; the even cases requires a little bit more work but follow the same principles.
Re-normalization
First I wish to do some change of variables to make typing a bit easier. If we replace $\rho = \sqrt{d}\pi r$, and $\tau = t / \sqrt{d}{\pi}$, the integral can be written as
$$ \frac{1}{(\sqrt{d}\pi)^{d-1}}\int_0^1 \rho^{d-2} \mathrm{d}\rho \int_0^\pi \sin(\tau\rho) \exp(i \nu \tau \rho \cos\theta) (\sin\theta)^{d-2} ~\mathrm{d}\theta $$
(here we also set $\nu = \sqrt{d}v$). Our goal reduces to studying the integral
$$ J(\tau;\nu,k) := \int_0^1 \rho^{2k-1} \mathrm{d}\rho \int_0^\pi \sin(\tau\rho) \exp(i \nu \tau \rho \cos\theta) (\sin\theta)^{2k-1} ~\mathrm{d}\theta $$
It is convenient to take another change of variables $z = \cos\theta$ to write
$$ J(\tau;\nu,k) = \int_0^1 \rho^{2k-1} \mathrm{d}\rho \int_{-1}^1 \sin(\tau\rho) \exp(i \nu \tau\rho z)(1 - z^2)^{k-1} ~\mathrm{d}z $$
$\nu = 0$
In this case the inner exponential drops, and we get no oscillation in $\theta$. Let $\gamma_k = \int_0^\pi (\sin\theta)^{2k-1} ~\mathrm{d}\theta > 0$, we have after integrating by parts
$$ J(\tau;0,k) = \gamma_k \int_0^1 \rho^{2k-1} \sin(\tau\rho) ~\mathrm{d}\rho = \frac{\gamma_k}{\tau} \left( - \cos(\tau) + \int_0^1 (2k-1) \rho^{2k-2} \cos(\tau\rho) ~\mathrm{d}\rho \right) $$
Integrating by parts again we find
$$ J(\tau;0,k) = -\frac{\gamma_k}{\tau}\cos(\tau) + O_{\tau\to\infty}(\tau^{-2}) $$
As $J(\tau;\nu,k)$ is continuous in $\nu$ for fixed $k$ and $\tau$, the asymptotics above shows that it is impossible to have any bound of the form
$$ |J(\tau;\nu,k)| \leq \frac{C_k}{\tau^{k}} $$
with $C_k$ dependent on $k$ but independent of $\nu$, when $k > 1$.
Angular integration
We may write
$$ \exp(i \nu \tau \rho z) = \frac{1}{i \nu \tau \rho} \frac{\mathrm{d}}{\mathrm{d}z}\exp(i \nu \tau \rho z) $$
Using that $(1-z^2)^{k-1}$ vanishes to order $k-1$ at $z = \pm 1$, we can integrate by parts $k$ times to get
$$ J(\tau;\nu,k) = \int_0^1 \rho^{k-1} \left( \frac{-1}{i \nu \tau }\right)^{k} \sin(\tau\rho) \mathrm{d}\rho \left[ \int_{-1}^1 \exp(i \nu\tau \rho z) \frac{\mathrm{d}^k}{\mathrm{d}z^k} (1-z^2)^{k-1} ~\mathrm{d}z - \left. \exp(i \nu \tau\rho z) \frac{\mathrm{d}^{k-1}}{\mathrm{d}z^{k-1}} (1-z^2)^{k-1} \right|_{z = -1}^{z = 1} \right]$$
This shows that there exists a constant $C_k$ such that
$$ | J(\tau;\nu,k) | \leq \frac{C_k}{\nu^k \tau^k} $$
which is the desired decay rate except for the fact that the rate depends on $\nu$.
Radial integration for small $\nu$
It turns out that the $\tau^{-1}$ decay at $\nu = 0$ can be extended to all $\nu$ small in a uniform way. We may write
$$ \sin(\tau\rho) = \frac{1}{2i} (e^{i\tau\rho} - e^{-i\tau\rho}) $$
and so
$$J(\tau;\nu,k) = J_+(\tau;\nu,k) - J_-(\tau;\nu,k)$$
where
$$J_\pm(\tau;\nu,k) = \int_0^1 \rho^{2k-1} \mathrm{d}\rho \int_{-1}^1 \exp(i \tau\rho (\nu z \pm 1)) (1-z^2)^{k-1} ~\mathrm{d} z $$
Integrating by parts in $\rho$ we find
$$J_\pm(\tau;\nu,k) = \int_{-1}^1 \frac{1}{i \tau (\nu z \pm 1)} (1 - z^2)^{k-1} \left[ \exp(i \tau(\nu z \pm 1)) - \int_0^1 (2k-1)\rho^{2k-2} \exp(i \tau\rho(\nu z \pm 1)) ~\mathrm{d}\rho \right] ~\mathrm{d} z $$
Using that $\nu \in [0,\frac12] \implies |\nu z \pm 1| \geq \frac12$ we conclude that there exists a constant $\tilde{C}_k$ dependent on $k$ but not on $\nu$ such that
$$\nu \in [0,\frac12] \implies |J_\pm(\tau;\nu,k)| \leq \frac{\tilde{C}_k }{\tau} $$
Result for $d = 3$ ($k = 1$)
Apply the result from the angular integration to $\nu \geq \frac12$, and the result from the radial integration to $\nu \leq \frac12$, we obtain that
$$ |J(\tau;\nu,1)| \leq \frac{2\max( \tilde{C}_1,C_1)}{\tau} $$
as hypothesized.
Additional radial cut-off
We remark again that the analysis at $\nu = 0$ already rules out the hypothesized bound when $k > 1$. The issue is firmly with the radial integral having an amplitude $\rho^{d-2}$ that does not vanish at the upper limit $\rho = 1$.
If $\rho^{d-2}$ is replaced by $\phi(\rho)$ which vanishes to order at least $\kappa$ (with $\kappa\in \mathbb{Z}_{\geq 1}$) at both $\rho = 0$ and $\rho = 1$, then we would be able to repeat the radial integration by parts $\kappa + 1$ times (exactly as the angular integration), and obtain that for $\nu \in [0,\frac12]$ the upper bound $\tilde{\tilde{C}}_{\kappa,k} / \tau^{\kappa + 1}$
