On the equality of derivatives of two functions.

For some $n$? Not even close. Let $n=1$, $f(x) =x$, and $g(x)=x+1$. Then $f'(x) = g'(x)$, but $f(\alpha) \neq g(\alpha)$, not at any point $\alpha$.

Close. You forget to include constant of integration:

$$f(x)=g(x)+a_0+a_1x+a_2x^2+\dots+a_{n-1}x^{n-1}$$

For arbitrary constants $a_k$.

For example:

$$f(x)=g(x)+a_0+a_1x+a_2x^2\tag1$$

$$f'(x)=g'(x)+\require{cancel}\cancelto0{\frac{da_0}{dx}}+\cancelto{a_1}{\frac{da_1x}{dx}}+\cancelto{2a_2x}{\frac{da_2x^2}{dx}}\\f'(x)=g'(x)+a_1+2a_2x$$

$$f''(x)=g''(x)+\cancelto0{\frac{da_1}{dx}}+\cancelto{2a_2}{\frac{d2a_2x}{dx}}$$

$$f''(x)=g''(x)+2a_2$$

$$f'''(x)=g'''(x)+\cancelto0{\frac{d2a_2}{dx}}$$

$$f'''(x)=g'''(x)$$

So the solution to that DE is given by $(1)$.

For fractional values of $n$, we run into fractional calculus, whereupon I cannot give you any simple answer.

I can say the following:

$$f(x)=g(x)\implies\frac{d^\alpha}{dx^\alpha}f(x)=\frac{d^\alpha}{dx^\alpha}g(x)$$

However,

$$\frac{d^\alpha}{dx^\alpha}f(x)=\frac{d^\alpha}{dx^\alpha}g(x)\cancel\implies f(x)=g(x)$$

If you are interested, you can try to look up some things on fractional calculus, which is concerned with fractional derivatives.

Here is the Wikipedia for fractional calculus, but do be warned that most sources ignore a fractional constant of integration or whatever due to its confusing nature.

You write: "What i know is that if the $n-th$ derivative of $f$ equals that of $g$ (where $n$ is an integer), then $f=g$, but i'm not sure if this holds for fractional $n$."

Not even this is true. Let $f(x) = 0$ and $g(x) = 1$. Then all positive integer derivatives of $f$ and $g$ are zero, but $f \neq g$ everywhere. Further, (though strictly not something you asked about,) that all positive integer derivatives of two functions agree does not mean that any of their positive non-integer derivatives agree. For positive non-integer $\alpha$, \begin{align} \frac{\mathrm{d}^\alpha}{\mathrm{d}x^\alpha} f(x) &= \frac{0 \cdot x ^{-\alpha}}{\Gamma(1-\alpha)} = 0 \\ \frac{\mathrm{d}^\alpha}{\mathrm{d}x^\alpha} g(x) &= \frac{1 \cdot x ^{-\alpha}}{\Gamma(1-\alpha)} \neq 0 \text{,} \end{align} where I have written "${} \neq 0$" because that function is never zero for any value of $x$. (For integer $\alpha$, we can recover the zero derivatives we have already mentioned by being careful taking our limits. When we are, the pole in $\Gamma$ dominates the algebraic numerator, giving zero.)

Now to your specific question. Fix a positive non-integer $n$ and let $f(x) = x$ and $g(x) = \frac{\Gamma(2)}{\Gamma(3)}\cdot\frac{\Gamma(3-n)}{\Gamma(2-n)} x^2$. Then \begin{align} \frac{\mathrm{d}^n}{\mathrm{d}x^n} f(x) &= \frac{\Gamma(2)}{\Gamma(2-n)} x^{1-n} \\ \frac{\mathrm{d}^n}{\mathrm{d}x^n} g(x) &= \frac{\Gamma(2)}{\Gamma(3)}\cdot\frac{\Gamma(3-n)}{\Gamma(2-n)} \cdot \frac{\Gamma(3)}{\Gamma(3-n)} x^{2-n} \\ &= \frac{\Gamma(2)}{\Gamma(2-n)} x^{2-n} \text{.} \end{align} And we can see that $f^{(n)}(1) = g^{(n)}(1) = \frac{\Gamma(2)}{\Gamma(2-n)}$. However, $f(1) = 1$ and $g(1) = \frac{\Gamma(2)}{\Gamma(3)}\cdot\frac{\Gamma(3-n)}{\Gamma(2-n)} = 1-\frac{n}{2}$, and this last expression is $1$ only for $n=0$, contradicting our choice of $n$ as positive non-integer.

We have shown that for any positive non-integer $n$, we can find two functions whose $n^\text{th}$ derivatives agree at a point but whose values do not agree at the same point. So just as for positive integer derivatives, two functions' derivatives agreeing at a point is insufficient to conclude that the two functions are equal at that point.

Short answer - no.

Possible correction: instead of "for some postive real n" try "for all postive integral n" (I'll not discuss non-integral derivatives here. That's a separate topic)

But still then, the answer is no.

consider: g(x)=f(x) + e^(-1/(x-a)^2)

f and g satisfy your conditions but are not equal except for X=a.