A partial answer to this question is as follows. Take any Brownian motion $B=(B_t)$ with a base $\mathcal B=(\Omega,\mathcal{F},\mathbb{F},\mathbb{P})$. Consider the
Karhunen–Loève expansion
$$B_t=\sqrt2\sum_{k=1}^\infty Z_k \frac{\sin((k-1/2)t)}{(k-1/2)\pi}
$$
for $t\in[0,1]$, where the $Z_k$'s are certain iid standard normal random variables, defined on the same probability space $\mathcal P:=(\Omega,\mathcal{F},\mathbb{P})$.
For $t\in[0,1]$, let
$$C_t=\sqrt2\sum_{k=1}^\infty Z_{2k-1} \frac{\sin((k-1/2)t)}{(k-1/2)\pi},\quad
D_t=\sqrt2\sum_{k=1}^\infty Z_{2k} \frac{\sin((k-1/2)t)}{(k-1/2)\pi}.
$$
Then $C$ and $D$ are independent Brownian motions on $[0,1]$, defined on the same probability space $\mathcal P$, but apparently not on the same base $\mathcal B$ -- which is why the answer is only partial.

Now it easy to use shifting in $t$ and gluing to construct two (or even countably many) independent Brownian motions on $[0,\infty)$, again on the same probability space. Here we need to partition the set of all natural numbers into countably many countable subsets (rather than into two subsets, of all odd natural numbers and all even natural numbers, as was done above). Here the term "countable" is of course used in the sense of "countably infinite".

On the other hand, take any Brownian motion $B=(B_t)$ and let $\mathcal B=(\Omega,\mathcal{F},\mathbb{F},\mathbb{P})$ be the (smallest) base generated by $B$, with $\Omega$ being the set of all paths of $B$. Then there is no Brownian motion $C$ on base $\mathcal B$ that is independent of $B$.

Indeed, otherwise the sigma-algebra $\mathcal G$ generated by $C$ would be a sub-sigma-algebra of $\mathcal F$ that is independent of $\mathcal F$. So, the sigma-algebra $\mathcal G$ would be independent of itself. That is only possible is all members of $\mathcal G$ have probability $0$ or $1$. But then $\mathcal G$ cannot be the sigma-algebra generated by a Brownian motion.

However, as noted by Robert Israel, there still might exist, on the same base, two independent Brownian motions $C_1$ and $C_2$.

So, the problem remains open.