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On the face of it this might just seem silly, but I wonder if I scribe a circle on a sphere will the ratio of the circumference of such a circle to its diameter would still be $\pi$. If so, is there a way I could prove it or disprove otherwise?

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    I'd like to know how are you inscribing a circle on a sphere? – 2017-01-05
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    @Shraddheya Shendre: I edited my question to replace inscribe by scribe. – 2017-01-05
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    Do you intend to discretize a circle on a sphere? [This question](http://math.stackexchange.com/questions/327699/discretize-a-circle-on-a-sphere-with-a-given-center-and-radius) might be of interest to you. – 2017-01-05

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I believe you're trying to stick to quantities measured on the surface of the sphere, right? In that case, you're basically working with spherical trigonometry. In this particular case, the 'radius' of the circle corresponds to a central (polar) angle of the sphere using the standard arc length formula: $$s = \theta R$$ if $R$ is the sphere's radius, $\theta$ is the angle, and $s$ is the circle's "radius".

Finding the arc length of the circle is a straightforward exercise in $3$-$d$ geometry (or integration of the line element) that gives: $$C = 2\pi R \sin\theta,$$ leading to: $$\begin{align} \frac{C}{D} & = \frac{C}{2s} \\ & = \pi \frac{\sin \theta}{\theta}. \\ & = \pi \frac{R \sin\left(\frac{s}{R}\right)}{s} \\ & \approx \pi \left( 1 + \frac{1}{3!} \left[\frac{s}{R}\right]^2 + \ldots\right) \end{align}$$

So, in a non-Euclidean geometry, if you define $\pi$ in terms of the ratio of circumference to diameter of a circle you get an answer that depends on the diameter of the circle. That's not how it's commonly used, though, since the value of $\pi$ is so useful outside of geometry. You can consider $\pi$ to be defined as: $$\pi \equiv \lim_{D\rightarrow0} \frac{C}{D},$$ in all sufficiently smooth spaces, or just define it in terms of one of the mathematical series.

If the Gaussian curvature is negative, but still constant everywhere, then the space is hyperbolic, and the $R$ in the circumference formulae becomes imaginary, giving: $$C = 2\pi R \sinh \left(\frac{s}{R}\right).$$

You can do similar exercises concerning the sum of the angles of a triangle to prove that they add up to some value that depends on the size of the triangle in these non-Euclidean spaces.

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Any circle satisfies the condition that the ratio of its circumference to its diameter is $\pi$, so yes.