3 added a proof in the case of manifolds of nonpositive curvature

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Your original question is false for the round sphere, since for the unit vector $v$ orthogonal to $\gamma'(0)$ you get $|J(t)|^2=\sin^2(t)$, while $\tan(t)>t$ for $t\in (0, \pi/2)$.

It follows immediately from the Taylor expansion for $|J(t)|$ that your inequality holds if curvature of $M$ is negative. The inequality is probably also true for manifolds of nonpositive curvature, but would require a bit more work.

Edit: Here is the proof of the inequality in the case of nonpositive sectional curvature. I will assume that $J'(0)$ is a unit vector orthogonal to $\gamma'(0)$ (since the proof in the case of the tangential Jacobi field $t\gamma'(t)$ is clear: you get the equality). Let $v(t)=|J(t)|^2$ and let $\tilde{v}=|\tilde{J}(t)|^2$, where $\tilde{J}$ is a Euclidean Jacobi field (so that $\tilde{J}(0)=0$, and $\tilde{J}'(0)$ is also a unit vector orthogonal to $\tilde{\gamma}'(0)$, where $\tilde{\gamma}$ is a Euclidean geodesic). Then you get the comparison inequality: $$v' \tilde{v} \ge v \tilde{v}'$$
for all $t$, see do Carmo's book "Riemannian Geometry", proof of Rauch comparison theorem, pages 216-217. You get: $\tilde{v}= t^2, \tilde{v}'=2t$ and the above comparison inequality becomes the inequality $$\frac{t}{2} v'\ge v$$ for all $t$, which is exactly the inequality that you are asking for (since for small $t$, $d(\gamma(t))=t$).

2 added 226 characters in body

This is false for the round sphere, since for the unit vector $v$ orthogonal to $\gamma'(0)$ you get $|J(t)|^2=\sin^2(t)$, while $\tan(t)>t$ for $t\in (0, \pi/2)$.

It follows from the Taylor expansion for $|J(t)|$ that your inequality holds if curvature of $M$ is negative. The inequality is probably also true for manifolds of nonpositive curvature, but would require a bit more work.

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This is false for the round sphere, since for the unit vector $v$ orthogonal to $\gamma'(0)$ you get $|J(t)|^2=\sin^2(t)$, while $\tan(t)>t$ for $t\in (0, \pi/2)$.