The answer in general is no, because there are simple counterexamples based on the Gauss-Bonnet theorem. On the other hand there is a sufficient condition which is of an easily checked combinatorial nature, based on W. Thurston's theory of train tracks.

We might as well assume that $\gamma_1,\gamma_2$ have minimal geometric intersection number, because that condition is necessary and sufficient for the existence of an ambient isotopy taking $\gamma_1 \cup \gamma_2$ to $\tilde\gamma_1 \cup \tilde\gamma_2$ no matter what the hyperbolic structure.

First the counterexample. For each complementary region $S$ of $\gamma_1 \cup \gamma_2$---meaning a component of $F-(\gamma_1 \cup \gamma_2)$---by taking the completion we may regard $S$ as the interior of a surface $\overline S$ with piecewise smooth boundary $\partial\overline S$, the smooth pieces alternating between segments of $\gamma_1$ and $\gamma_2$. The orientation on each smooth piece obtained by restricting the $\gamma_1$ or $\gamma_2$ orientation may or may not agree with the boundary orientation on $\partial \overline S$. The point of saying "completion" rather than simply "closure" is that $S$ may touch itself from both sides of some smooth piece, or from opposite angles at some intersection point.

The counterexample is chosen so that for some $S$ its completion $\overline S$ is a disc with 4 sides, and the restricted orientations on $\partial\overline S$ all agree with the boundary orientation. It is not hard, with some quick sketches, to construct $\gamma_1,\gamma_2$ having such a complementary region $S$. Assuming that $\gamma_1,\gamma_2$ are geodesic, the surface $\overline S$ may therefore be regarded as a finite quadrilateral in $\mathbb{H}^2$. Assuming there existed a hyperbolic metric with $\gamma_1,\gamma_2$ geodesic and with acute angles $<\frac{\pi}{2}$ as said, as one goes around $\partial \overline S$ we see four exterior angles each $<\frac{\pi}{2}$, whose sum is $<2\pi$. However, by the Gauss-Bonnet theorem the sum of the exterior angles of an $\mathbb{H}^2$ polygon equals $2\pi$ plus the area, so that sum is $>2 \pi$.

Next the sufficient condition, which has two clauses.

First, form a "pre train track" $\tau$ from $\gamma_1 \cup \gamma_2$ by smoothing $\gamma_1 \cup \gamma_2$ at each intersection point $p_i \in \gamma_1 \cap \gamma_2$ so that $\gamma_1,\gamma_2$ are tangent at $p_i$ and they induce the same orientation on their common tangent line at $p_i$. The intuitive combinatorial effect of this smoothing is that the angles you want to be "acute" are taken to a limit of a zero angle, and their supplementary "obtuse" angles are taken to a limit of a straight angle.

The first clause of the sufficient condition is that $\tau$ is a true "train track". By definition this means that for each complementary region $S$, the quantity
$$i(S) = \chi(S)\, -\, \frac{1}{2} \, \# \text{(nonsmooth boundary points of $S$)}
$$
is negative. This first clause rules out the counterexample above, because in that counterexample the smoothing has the effect that the disc $S$ has no nonsmooth boundary points and so $i(S)=\chi(S)=1$ which is positive.

Next there is a theorem you can find in the book of Harer and Penner, which they learned from Nat Kuhn, and which goes back to Thurston's early lectures on train tracks and hyperbolic structures. The theorem says that the following properties of $\tau$ are equivalent:

For each $\epsilon>0$ and $L>0$ there exists a hyperbolic structure on the surface in which each branch of $\tau$ is a geodesic arc of length $\ge L$ and at each point $p_i \in \gamma_1 \cup \gamma_2$ the angle made by a train path of $\tau$ through $p_i$ differs from a straight angle by $\le \epsilon$.

The train track $\tau$ is transversely recurrent.

Item 1 implies what you want, in fact is quite a bit stronger because it guarantees that the acute angles at the points $p_i$ can be made as small as possible. Item 2 is a combinatorial condition which can be easily checked in practice.

So the second clause of the sufficient condition is that $\tau$ is transversely recurrent.