Exponential of approximate quadratic variation of Brownian motion Let $X_t$ be a Brownian motion or a Brownian Bridge on a (\edit: compact) Riemannian manifold. Let $T>0$ be given. 
The question is: Does there exists a constant $C>0$ such that for all partitions $0 = \tau_0 < \tau_1 < \dots < \tau_N \leq T$, we have
$$ \mathbb{E}\left[ \exp \left(\sum_{j=1}^N d(X_{\tau_{j-1}}, X_{\tau_j})^2 \right) \right] \leq C~~~~~~?$$
I can prove so far that for all $p \in [1, \infty)$, there exists a constant $C_p$ such that
$$\mathbb{E}\left[ \left(\sum_{j=1}^N d(X_{\tau_{j-1}}, X_{\tau_j})^2 \right)^p \right] \leq C_p,$$
but I cannot control the constants $C_p$ so that the "brute force proof" using the exponential series fails.
 A: Yes, there is a bound like that for $T \le \mathrm{const}$. I'll do the case of Brownian motion, since the Brownian bridge reduces to it.
The proof consists of two stages: proving the bound in hyperbolic space and reducing the general case to it.
In order to redue the general case to the constant curvature case we can use the comparison theorem for the Laplacian of the distance function (Theorem 1.1 in cvgmt.sns.it/media/doc/paper/2113/HessBV.pdf). Namely:

Theorem: Fix a point $p$ and take the distance function $d_p(\cdot) := d(p, \cdot)$. Then a curvature bound
  $$\operatorname{Ric} \ge (n-1) K$$
  for some constant $K$ implies a bound
  $$\Delta d_p(r, \theta) \le \Delta^K d^K(r)$$
  outside of the cut locus. Here $\Delta^K$ and $d^K$ refers to the Laplacian and distance function of the simply connected constant curvature $K$ space.

For our purposes we won't need to touch the cut locus, so this most naive version suffices.
Recall that the distance process $d_p(X_t)$ (resp. its constant curvature counterpart $d^K(X_t^K)$) is a semimartingale with drift $\frac{1}{2} \Delta d_p(X_t)$ (resp. $\Delta^K d^K(X_t^K)$) and quadratic variation $t$, so they can be represented as solutions to SDEs
$$d (d_p(X_t)) = \frac{1}{2} \Delta d_p(X_t) dt + dB_t$$
$$d (d^K(X_t^K)) = \frac{1}{2} \Delta^K d^K(X_t^K) dt + dB_t$$
where $B$ is a standard $1$-dimensional Brownian motion. By the Laplacian comparison theorem we have an inequality between the drift terms, and I intentionally used the same driving noise $dB$ in both SDEs. This way the processes become coupled so that

$d_p(X_0) \le d_p^K(X_0^K)$ implies $d_p(X_t) \le d_p^K(X_t^K)$ for all $t$ at least until $X_t$ meets the cut locus.

(see e.g. the proof of Theorem 20.5 in Kallenberg's "Foundations of modern probability")
So let's start $X_t$ at the point $p$ and consider the stopped process $X_{t \wedge \theta}$, $\theta := \inf\{t : d_p(X_t) = R\}$, where $R$ is the injectivity radius of our manifold. Similarly, take $\theta^K := \inf\{t : d^K(X_t^K) = R\}$. From the above reasoning, $d_p(X_{t \wedge \theta}) \le d^K(X^K_{t \wedge \theta^K})$ for all $t$, so
$$\mathsf{E} \exp (d(X_0, X_{t\wedge\theta}))^2 \le \mathsf{E} \exp (d^K(X_0^K, X^K_{t\wedge\theta^K}))^2,$$
$$\theta \ge \theta^K$$
In order to get rid of this $\theta$ stopping, just use the trivial inequality
$$\mathsf{E} \exp (d(p, X_t))^2 \le \mathsf{E} \exp (d(p, X_{t \wedge \theta}))^2 + \exp D^2 \cdot \mathsf{P} \{\theta \le t\},$$
where $D$ is the diameter of our manifold.
Finally, this gives:
$$\mathsf{E} \exp (d(p, X_t))^2 \le \mathsf{E} \exp(d^K(X_t^K))^2 + \exp D^2 \cdot \mathsf{P}\{\theta_K \le t\}$$
Denote the right-hand side by $F(t)$.
Now note that whenever we have a sequence of times $0 = \tau_0 < \dots \le \tau_N$ we can use the Markov property of the BM to get the same bound conditionally on $X_{\tau_{N-1}}$, then on $X_{\tau_{N-2}}$, etc.:
$$\mathsf{E} \prod_{k < N} \exp (d(X_{\tau_k}, X_{\tau_{k+1}}))^2 \le \dots \le F(\tau_1 - \tau_0) \dots F(\tau_N - \tau_{N-1})$$
Now, in order to deal with the constant curvature case one can use the same approach as above: use a bound like $\Delta^K d^K(r) \le \frac{n-1}{r} + O(1)$ in order to dominate $d^K(X_t^K)$ by a Bessel($n$) process with constant drift, to reduce everything to the trivial case of Brownian motion in $\mathbb{R}^n$.
Similarly, the Brownian bridge case reduces to Brownian motion. Indeed, Brownian bridge on a time interval $[0,T/2]$ is just a Brownian motion with bounded drift, and $[T/2, T]$ it's a time-reversed Brownian motion with bounded drift.
