Let $W$ be a one dimensional standard Brownian motion, and $\sigma: [0, \infty) \to \mathbb R$ a Borel function with $c < \sigma < C$ for some constants $c, C > 0$.
Does there exist some $M > 0$ such that, conditional on $W_t > -1$ for all $0 \leq t \leq 1$ and $W_1 > M$,
$\int_0^1 \sigma(t) \, dW_t \geq 0$, almost surely?
Remark: The statement would be true if the integrator had uniformly bounded variation, but I suspect it does not hold for Brownian motion which has unbounded variation.
No such $M$ exists for the following $\sigma$. Partition $[0,1]$ into countably many intervals, with endpoints $t_0=0,t_1,t_2,...$ and let $\sigma$ take value 1 on the odd ones and value 2 on the even ones. Given any $M$, the following event $A_M$ has positive probability:
$A_M$ requires that the increments $W_{t_k}-W_{t_{k-1}}$ are in $(9,10)$ for the first $M$ odd values of $k$, and in $(-6,-7)$ for the first $M$ even values of $k$, with $W_t-W_{t_{k-1}}>-1$ for $t \in [t_{k-1},t_k]$ when $0<k<2M$.
Then given $A_M$, we have $W_{t_{2M}}>2M$, so the event $W_1>M$ holds with high probability, yet on $A_M$, the stochastic integral considered is $<-2M$ if you integrate up to $t_{2M}$, and the integral over $[t_{2M},1]$ is likely to be $<2M$.
